datavorous/entrance-exam-dataset huggingface dataset:
The equations of the tangents of the ellipse $9 x^2+16 y^2=144$ which passes through the point $(2,3)$ is

Options:
A $y=3, x+y=5$
B $y=-3, x-y=5$
C $y=4, x+y=3$
D $y=-4, x-y=3$

The correct answer is:
$y=3, x+y=5$

Solution:
the slope form of tangent to the given ellipse is:
$y=m x \pm 16 m^2+9$
Also, it passes through $(2,3)$
$\begin{aligned}
& \therefore 3=2 \mathrm{~m}+\sqrt{16 \mathrm{~m}^2+9} \\
& \Rightarrow(3-2 \mathrm{~m})^2=16 \mathrm{~m}^2+9 \\
& \Rightarrow 12 \mathrm{~m}^2+12 \mathrm{~m}=0 \\
& \Rightarrow \mathrm{m}=0,-1
\end{aligned}$
Hence, the equation of the tangents are:
$y=3 \text { and } y=-x+5$

Topic: Mathematics, Ellipse, JEE Main
The equations of the tangents to the circle $x^2+y^2-6 x+4 y=12$ which are parallel to the straight line $4 x+3 y+5=0$, are

Options:
A $3 x-4 y-19=0,3 x-4 y+31=0$
B $4 x+3 y-19=0,4 x+3 y+31=0$
C $4 x+3 y+19=0,4 x+3 y-31=0$
D $3 x-4 y+19=0,3 x-4 y+31=0$

The correct answer is:
$4 x+3 y+19=0,4 x+3 y-31=0$

Solution:
Let equation of tangent be $4 x+3 y+k=0$, then $\sqrt{9+4+12}=\left|\frac{4(3)+3(-2)+k}{\sqrt{16+9}}\right|$ $\Rightarrow 6+k= \pm 25 \Rightarrow k=19$ and -31
Hence the tangents are $4 x+3 y+19=0$ and $4 x+3 y-31=0$.

Topic: Mathematics, Circle, JEE Main
The equations of three sides of a triangle are \( x=2, y+1=0 \) and \( x+2 y=4 \). The coordinates of the circumcentre of the triangle are \( (\lambda, \mu) \), then \( \lambda+\mu \) is

Options:

The correct answer is:
4

Solution:
One side of the triangle is parallel to the Y-axis and another side is parallel to the X-axis. So, the triangle is a right-angled triangle. Hence, the mid-point of the hypotenuse is the circumcentre. Solving, x=2, x+2 y=4, we get one end of the hypotenuse and solving y+1=0, x+2 y=4, we get the other end. Their coordinates are A(2,1) and B(6-1)

∴Circumcentre =mid-point of AB=2+62,1-12=4,0=λ,μ

⇒λ+μ=4

Topic: Mathematics, Straight Lines, JEE Main
The equations of three waves are given by y1=A0sinkx−ωt, y2=32A0sinkx−ωt+ϕ and y3=4A0coskx−ωt. Theses waves are in the same direction and are superimposed. The phase difference between the resultant-wave and the first wave is π4 and ϕ=πn≤π2, then what is the value of n?

Options:

The correct answer is:
12.00

Solution:
tanπ4=BCAC=A0(4+32sinϕ)A0(1+32cosϕ)

⇒cosϕ-sinϕ=12
Squaring both sides, ⇒cos2ϕ+sin2ϕ-2cosϕsinϕ=12
⇒2sinϕcosϕ=12
⇒sin2ϕ=12
⇒ϕ=12sin-112=π12

Topic: Physics, Waves and Sound, JEE Main
The equations of two sides $\mathrm{AB}$ and $\mathrm{AC}$ of a triangle $\mathrm{ABC}$ are $4 x+y=14$ and $3 x-2 y=5$, respectively. The point $\left(2,-\frac{4}{3}\right)$ divides the third side $\mathrm{BC}$ internally in the ratio $2: 1$. the equation of the side $\mathrm{BC}$ is

Options:
A $x+3 y+2=0$
B $x-6 y-10=0$
C $x-3 y-6=0$
D $x+6 y+6=0$

The correct answer is:
$x+3 y+2=0$

Solution:
$\begin{aligned}
& \frac{2 \mathrm{x}_2+\mathrm{x}_1}{3}=2, \frac{2\left(\frac{3 \mathrm{x}_2-5}{2}\right)+\left(14-4 \mathrm{x}_1\right)}{3}=\frac{-4}{3} \\
& 2 \mathrm{x}_2+\mathrm{x}_1=6,3 \mathrm{x}_2-4 \mathrm{x}_1=-13 \\
& \mathrm{x}_2=1, \mathrm{x}_1=4
\end{aligned}$
So, $\mathrm{C}(1,-1), \mathrm{B}(4,-2)$
$\mathrm{m}=\frac{-1}{3}$
Equation of $\mathrm{BC}: \mathrm{y}+1=\frac{-1}{3}(\mathrm{x}-1)$
$\begin{aligned} & 3 y+3=-x+1 \\ & x+3 y+2=0\end{aligned}$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2024 (08 Apr Shift 1)
The equations of two sides of a variable triangle are x=0 and y=3, and its third side is a tangent to the parabola y2=6x. The locus of its circumcentre is :

Options:
A 4y2-18y-3x-18=0
B 4y2+18y+3x+18=0
C 4y2-18y+3x+18=0
D 4y2-18y-3x+18=0

The correct answer is:
4y2-18y+3x+18=0

Solution:
Given,

Equation of parabola is y2=6x,

Also the general equation of a parabola is y2=4ax

On comparing, we get

⇒4a=6⇒a=32

Tangent on parabola y2=4ax is given by y=mx+am,

So, y=mx+32 m; (m≠0)

Now using the midpoint formula we get,

h=6 m-34 m2, k=6 m+34 m

Now, eliminating m and we get 3h=2-2k2+9k-9

⇒3h=-4k2+18k-18

⇒4k2-18k+3h+18=0

∴ The locus of its circumcentre is

4y2-18y+3x+18=0

Topic: Mathematics, Parabola, JEE Main, JEE Main 2023 (25 Jan Shift 2)
The equations of two waves are given by :

y1=5sin2πx-vt cm

y2=3sin2πx-vt+1.5 cm

These waves are simultaneously passing through a string. The amplitude of the resulting wave is :

Options:
A 2 cm
B 4 cm
C 5.8 cm
D 8 cm

The correct answer is:
2 cm

Solution:
Given equations can be written as,

y1=5sin2πx-2πvt cm and

y2=3sin2πx-2πvt+3π cm

From the given equations we can conclude that the phase difference between the waves is 3π i.e. odd multiple of π. Therefore, destructive interference will happen.

A=A1-A2=2 cm

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2022 (24 Jun Shift 1)
The equations $x=\frac{t}{4}, y=\frac{t^2}{4}$ represents

Options:
A A circle
B A parabola
C An ellipse
D A hyperbola

The correct answer is:
A parabola

Solution:
Eliminating $t$, we get
$16 x^2=4 y \Rightarrow x^2=\frac{1}{4} y$, which is a parabola.

Topic: Mathematics, Parabola, JEE Main
The equilibrium constant at 298 K for a reaction A+B⇌C+D is 100. If the initial concentration of all the four species were1 M each, then the equilibrium concentration of D (in mol L-1) will be:

Options:
A 1.818
B 1.182
C 0.182
D 0.818

The correct answer is:
1.818

Solution:
K at 298 K for the reaction
A+B ⇌C+D is 100.
Given: The initial concentration of all four species is 1 M.
A+B⇌C+Dt=01111at equilbrium1-x1-x1+x1+x
K=1+x 1+x1-x 1-x=100
⇒ 1+x1-x=10
⇒ x=911=0.818
⇒ D=1+x=1+0.818=1.818

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2016 (03 Apr)
The equilibrium constant at $850 \mathrm{~K}$ for the reaction $\mathrm{N}_2(g)+\mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{NO}(g)$ is 0.5625 . The equilibrium concentration of $\mathrm{NO}(g)$ is $3.0 \times 10^{-3} \mathrm{M}$. If the equilibrium concentrations of $\mathrm{N}_2(g)$ and $\mathrm{O}_2(g)$ are equal, the concentrations of $\mathrm{N}_2(g)$ in $M$ is

Options:
A $4.0 \times 10^{-3}$
B $4.0 \times 10^{-2}$
C $1.6 \times 10^{-3}$
D $3.0 \times 10^{-3}$

The correct answer is:
$4.0 \times 10^{-3}$

Solution:
Given,
$\mathrm{N}_2(g)+\mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{NO}(g)$
i.e. equilibrium constant, $K_C=0.5625$
Equilibrium concentration of $[\mathrm{NO}]=3 \times 10^{-3} \mathrm{M}$
$\begin{array}{rlrl}
& & K_C & =\frac{\left[\mathrm{NO}^2\right.}{\left[\mathrm{N}_2\right]\left[\mathrm{O}_2\right]} \\
& \text {But, } & {\left[\mathrm{N}_2\right]} & =\left[\mathrm{O}_2\right] \quad \text{(Given)} \\
& \therefore & K_C & =\frac{[\mathrm{NO}]^2}{\left[\mathrm{~N}_2\right]^2} \\
& & & \\
\text {or, } & 0.5625 & =\frac{\left[3 \times 10^{-3}\right]^2}{\left[\mathrm{~N}_2\right]^2} \\
& \text {or, } & {\left[\mathrm{N}_2\right]^2} & =\frac{\left[3 \times 10^{-3}\right]^2}{0.5625}=\frac{9 \times 10^{-6}}{0.5625} \\
& \text {or, } & {\left[\mathrm{N}_2\right]^2} & =16 \times 10^{-6} \\
\text {or, } & {\left[\mathrm{N}_2\right]} & =4 \times 10^{-3}
\end{array}$
Hence, option (a) is the correct answer.

Topic: Chemistry, Chemical Equilibrium, JEE Main
The equilibrium constant for a reaction A+B⇋C+D is 1×10-2 at 298 K and is 2 at 273 K. The chemical process resulting in the formation of C and D is:

Options:
A Exothermic
B Endothermic
C Unpredictable
D There is no relationship between ΔH and K

The correct answer is:
Exothermic

Solution:
log K 2 K 1 = ΔH 2.303R 1 T 1 − 1 T 2
T 1 > T 2 â&â K 2 > K 1
ΔH =−Ve

Topic: Chemistry, Chemical Equilibrium, JEE Main
The equilibrium constant for a reaction is \( 10 \). What will be the magnitude value of \( \Delta \mathrm{G}^{\circ} \) ?
\( \mathrm{R}=8.31 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}, \mathrm{~T}=314 \mathrm{~K} .\left(\right. \) Approximate integer value in \( \left.\mathrm{KJ} \mathrm{mol}^{-1}\right) \)

Options:

The correct answer is:
6

Solution:
Given:

Keq=10R=8.31 JK-1 mol-1T=314 K

As we know, the relation between the standard Gibbs free energy change and equilibrium constant is given by:

∆G0=-2.303 RT log Keq∆G0=-2.303 8.31314 log 10∆G0≃-6

So, the magnitude will be 6.

Topic: Chemistry, Thermodynamics (C), JEE Main
The equilibrium constant for :

CN -+ CH 3 COOH â HCN + CH 3 COO - is :

(GivenâpKb for CN-=4.69âandâpKb âforââCH3COO-=2.25)

Options:
A 3.7 Ã 1 0 4
B 2.8 × 1 0 - 5
C 1.97 × 1 0 4
D 0.5 × 1 0 - 5

The correct answer is:
3.7 × 1 0 4

Solution:
K eq. = HCN CH 3 COO - CN - CH 3 COOH ;
K CH 3 COOH = CH 3 COO - H + CH 3 COOH ;
K HCN = H + CN - HCN
pK b  of  CN − =4.69 ∴ pK HCN =9.31 ;  K HCN =4.9 × 10 −10
pK CH 3 COOH = 4.75 ; K CH 3 COOH = 1.77 Ã 1 0 - 5
â´ K eq. = K CH 3 COOH K HCN = 1.77 Ã 1 0 - 5 4.9 Ã 1 0 - 1 0 = 3.7 Ã 1 0 4

Topic: Chemistry, Ionic Equilibrium, JEE Main
The equilibrium constant for mutarotation $\alpha$-D Glucose $\rightleftharpoons \beta$-D Glucose is 1.8 . What percentage of $\alpha$ form remains at equilibrium?

Options:
A $35.7$
B $64.3$
C $55.6$
D $44.4$

The correct answer is:
$35.7$

Solution:
$\alpha$-D glucose $\rightleftharpoons \beta$-D glucose
$\begin{array}{lcc}\text { Initial } & 1 & 0 \\ \text { At equilibrium } & 1-\alpha & \alpha\end{array}$
$K=\frac{\alpha}{1-\alpha}=1.8$
Solving $\alpha=0.642$
$1-\alpha=0.358$
Percent of $\alpha$-D glucose remaining at equilibrium $=35.8 \%$.

Topic: Chemistry, Chemical Equilibrium, AIIMS, AIIMS 2009
The equilibrium constant for the disproportionation reaction $2 \mathrm{Cu}^{+}(\mathrm{aq}) \longrightarrow \mathrm{Cu}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{ag})$ at $25^{\circ} \mathrm{C}$

$\left(\mathrm{E}^{\circ} \mathrm{Cu}^{+} / \mathrm{Cu}=0.52 \mathrm{~V}, \mathrm{E}^{\circ} \mathrm{Cu}^{2+} / \mathrm{Cu}=0.16 \mathrm{~V}\right)$ is

Options:
A $6 \times 10^{4}$
B $6 \times 10^{6}$
C $1.2 \times 10^{6}$
D $1.2 \times 10^{-6}$

The correct answer is:
$1.2 \times 10^{6}$

Solution:
The reaction

$$

\begin{array}{l}

2 \mathrm{Cu}^{+}(\mathrm{aq}) \longrightarrow \mathrm{Cu}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq}) \\

\mathrm{E}_{\mathrm{cell}}=\mathrm{E}_{\mathrm{cell}}^{\circ}-\frac{0.0592}{1} \log \frac{\left[\mathrm{Cu}^{2+}\right]}{\left[\mathrm{Cu}^{+}\right]^{2}}

\end{array}

$$

At equilibrium $\mathrm{E}_{\mathrm{cell}}=0$

$$

\therefore \quad \mathrm{E}_{\text {cell }}^{\circ}=0.0592 \log \mathrm{K}_{\mathrm{c}}

$$

or, $\quad \log \mathrm{K}_{\mathrm{c}}=\frac{0.52-0.16}{0.0592}$

$\therefore \mathrm{K}_{\mathrm{c}}=1.2 \times 10^{6}$

Topic: Chemistry, Electrochemistry, JEE Main
The equilibrium constant for the following reactions are given at $25^{\circ} \mathrm{C}$
$\begin{array}{l}
2 \mathrm{~A} \rightleftharpoons \mathrm{B}+\mathrm{C}, \mathrm{K}_{1}=1.0 \\
2 \mathrm{~B} \rightleftharpoons \mathrm{C}+\mathrm{D}, \mathrm{K}_{2}=16 \\
2 \mathrm{C}+\mathrm{D} \rightleftharpoons 2 \mathrm{P}, \mathrm{K}_{3}=25
\end{array}$
The equilibrium constant for the reaction $\mathrm{P} \rightleftharpoons \mathrm{A}+\frac{1}{2} \mathrm{~B}$ at $25^{\circ} \mathrm{C}$ is

Options:
A $\frac{1}{20}$
B 20
C $\frac{1}{42}$
D 21

The correct answer is:
$\frac{1}{20}$

Solution:
Hint:
We can manipulate the given equations as follows
$\frac{1}{2} \mathrm{~B}+\frac{1}{2} \mathrm{C} \rightleftharpoons \mathrm{A}, \quad \mathrm{K}_{1}^{\prime}=1$
$\frac{1}{2} \mathrm{C}+\frac{1}{2} \mathrm{D} \rightleftharpoons \mathrm{B}, \quad \mathrm{K}_{2}^{\prime}=\frac{1}{\sqrt{16}}$
$\mathrm{P} \rightleftharpoons \mathrm{C}+\frac{1}{2} \mathrm{D}, \quad \mathrm{K}_{\mathrm{3}}^{\prime}=\frac{1}{\sqrt{25}}$
$\mathrm{P} \rightleftharpoons \mathrm{A}+\frac{1}{2} \mathrm{~B}, \quad \mathrm{~K}_{\text {final }}=1 \times \frac{1}{4} \times \frac{1}{5}=\frac{1}{20}$

Topic: Chemistry, Chemical Equilibrium, JEE Main
The equilibrium constant for the given reaction is 100 .
$\mathrm{N}_2(g)+2 \mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{NO}_2(g)$
What is the equilibrium constant for the reaction given below?
$\mathrm{NO}_2(g) \rightleftharpoons \frac{1}{2} \mathrm{~N}_2(g)+\mathrm{O}_2(g)$

Options:
A 10
B 1
C 0.1
D 0.01

The correct answer is:
0.1

Solution:
$\begin{gathered}\mathrm{N}_2+2 \mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO}_2 \\ K_1=\frac{\left[\mathrm{NO}_2\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{O}_2\right]^2}\end{gathered}$
or $\quad 100=\frac{\left[\mathrm{NO}_2\right]^2}{\left[\mathrm{~N}_2\right]\left[\mathrm{O}_2\right]^2} \quad \ldots$ (i)
Again, $\begin{aligned} {\left[\mathrm{NO}_2\right] } & \rightleftharpoons \frac{1}{2} \mathrm{~N}_2+\mathrm{O}_2 \\ K_2 & =\frac{\left[\mathrm{N}_2\right]^{1 / 2}\left[\mathrm{O}_2\right]}{\left[\mathrm{NO}_2\right]}\end{aligned}$
Eqs. (i) $\times$ (ii), we get
$\begin{gathered}
100 \times K_2^2=1 \\
\text { or } \quad K_2^2=\frac{1}{100} \quad \text { or } \quad K_2=\frac{1}{10}=0.1
\end{gathered}$

Topic: Chemistry, Chemical Equilibrium, JEE Main
The equilibrium constant for the given reaction is approximately 10â3

HPO 4 2â aq + HCO 3 â aq â H 2 PO 4 â aq + CO 3 2â aq

Which is strongest conjugate base in the given reaction?

Options:
A HPO42-aq
B HCO3-aq
C H2PO4-aq
D CO32-aq

The correct answer is:
CO32-aq

Solution:
CO 3 2− + H 2 PO 4 − ⇌ HPO 4 2− + HCO 3 2− Kc= 10 3
Since the value of K c for this reaction is greater than 1, it means this reaction is favourable in forward direction. Therefore, CO32- is the stronger conjugate base than HPO42-

Topic: Chemistry, Ionic Equilibrium, JEE Main
The equilibrium constant for the reaction
H2Og+COg⇌H2g+CO2g
is 0.44 at 1260 K. The equilibrium constant for the reaction
2H2g+2CO2g⇌2COg+2H2Og
at 1260 K is equal to

Options:
A 0.44
B 0.88
C 5.16
D 126

The correct answer is:
5.16

Solution:
2H2(g)+2CO(g)⇌2CO(g)+2H2O(g) is obtained by reversing the reaction
H2O(g)+CO(g)⇌H2(g)+CO2(g) and then multiplying it by the factor '2'. Hence the equilibrium constant ( K' ) for this reaction is given by
K'= ( 1 K ) 2 = ( 1 0.44 ) 2 =5.16

Topic: Chemistry, Chemical Equilibrium, JEE Main
The equilibrium constant for the reaction
\( \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 N O(g) \) is \( 4 \times 10^{-4} \) at \( 2000 \mathrm{~K} \). In presence of a catalyst the equilibrium is
attained ten times faster. Therefore the equilibrium constant in presence of catalyst of \( 2000 \mathrm{~K} \)
is

Options:
A \( 4 \times 10^{-3} \)
B \( 4 \times 10^{-2} \)
C \( 4 \times 10^{-3} \)
D \( 4 \times 10^{-4} \)

The correct answer is:
\( 4 \times 10^{-4} \)

Solution:
Since, equilibrium constants \( (K C) \) is not affected by addition of catalyst. Therefore, the equilibrium constant value
remains \( 4 \times 10^{-4} \).

Topic: Chemistry, Polymers, KCET, KCET 2017
The equilibrium constant for the reaction $\mathrm{N}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})$ at temperature $\mathrm{T}$ is $4 \times 10^{-4}$. The value of $\mathrm{Kc}$ for the reaction $\mathrm{NO}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{~N}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})$ at the same temperature is

Options:
A
$2.5 \times 10^2$

B
$0.02$

C
$4 \times 10^{-4}$

D
50

The correct answer is:
50

Solution:
As per property of equilibria reverse the equation \& divide it by 2

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2004
The equilibrium constant for the reaction $\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})$ is $4 \times 10^{-4}$ at $2000 \mathrm{~K}$. In presence of a catalyst the equilibrium is attained 10 times faster. Therefore, the equilibrium constant, in presence of the catalyst at $2000 \mathrm{~K}$ is

Options:
A $4 \times 10^{-4}$
B $4 \times 10^{-3}$
C $4 \times 10^{-5}$
D $2.5 \times 10^{-4}$

The correct answer is:
$4 \times 10^{-4}$

Solution:
As per Van't Hoff Equation
$\frac{\mathrm{d}\left(\ln \mathrm{K}_{\mathrm{eq}}\right)}{\mathrm{dT}}=\frac{\Delta \mathrm{H}_{\mathrm{T}}}{\mathrm{RT}^{2}}$
As catalyst don't affect enthalpy of reaction $\left(\Delta \mathrm{H}_{\mathrm{r}}\right)$ and Temperature is constant. Therefore, $\mathrm{K}_{\mathrm{eq}}$ (equilibrium constant) will remain constant i.e. $4 \times 10^{-4}$.

Topic: Chemistry, Chemical Equilibrium, JEE Main
The equilibrium constant for the reaction $\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})$ is $4 \times 10^{-4}$ at $2000 \mathrm{~K}$. In presence of a catalyst the equilibrium is attained 10 times faster. Therefore, the equilibrium constant, in presence of the catalyst at $2000 \mathrm{~K}$ is

Options:
A $4 \times 10^{-4}$
B $4 \times 10^{-3}$
C $4 \times 10^{-5}$
D $2.5 \times 10^{-4}$

The correct answer is:
$4 \times 10^{-4}$

Solution:
As per Van't Hoff Equation
$\frac{\mathrm{d}\left(\ln \mathrm{K}_{\mathrm{eq}}\right)}{\mathrm{dT}}=\frac{\Delta \mathrm{H}_{\mathrm{T}}}{\mathrm{RT}^{2}}$
As catalyst don't affect enthalpy of reaction $\left(\Delta \mathrm{H}_{\mathrm{r}}\right)$ and Temperature is constant. Therefore, $\mathrm{K}_{\mathrm{eq}}$ (equilibrium constant) will remain constant i.e. $4 \times 10^{-4}$.

Topic: Chemistry, Chemical Equilibrium, WBJEE, WBJEE 2021
The equilibrium constant for the reaction
$\mathrm{SO}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \rightleftharpoons \mathrm{SO}_3(g)$ is $5 \times 10^{-2}$ atm.
The equilibrium constant of the reaction
$2 \mathrm{SO}_3(g) \rightleftharpoons 2 \mathrm{SO}_2(g)+\mathrm{O}_2(g)$ would be

Options:
A $100 \mathrm{~atm}$
B $200 \mathrm{~atm}$
C $4 \times 10^2 \mathrm{~atm}$
D $6.25 \times 10^4 \mathrm{~atm}$

The correct answer is:
$4 \times 10^2 \mathrm{~atm}$

Solution:
$\begin{aligned} \mathrm{SO}_2+\frac{1}{2} \mathrm{O}_2 & \rightleftharpoons \mathrm{SO}_3 \\ K_1 & =\frac{\left[\mathrm{SO}_3\right]}{\left[\mathrm{SO}_2\right]\left[\mathrm{O}_2\right]^{1 / 2}} \\ 2 \mathrm{SO}_3 & \rightleftharpoons 2 \mathrm{SO}_2+\mathrm{O}_2 \\ \mathrm{~K}_2 & =\frac{\left[\mathrm{SO}_2\right]^2\left[\mathrm{O}_2\right]}{\left[\mathrm{SO}_3\right]^2}\end{aligned}$
From Eqs. (i) and (ii)
$\begin{aligned} K_2 & =\frac{1}{K_1^2} \\ K_2 & =\frac{1}{\left(5 \times 10^{-2}\right)^2}=\frac{1}{25 \times 10^{-4}} \\ & =\frac{100 \times 10^2}{25} \\ & =4 \times 10^2 \mathrm{~atm}\end{aligned}$

Topic: Chemistry, Chemical Equilibrium, JEE Main
The equilibrium constant for the reaction
$$
\mathrm{SO}_3(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})
$$
is $\mathrm{K}_{\mathrm{c}}=4.9 \times 10^{-2}$. The value of $\mathrm{K}_{\mathrm{c}}$ for the reaction
$$
2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{~g})
$$
will be

Options:
A
416

B
$2.40 \times 10^{-3}$

C
$9.8 \times 10^{-2}$

D
$4.9 \times 10^{-2}$

The correct answer is:
416

Solution:
$\begin{aligned} \mathrm{K}_{\mathrm{c}}^{\prime} & =\left(\frac{1}{4.9 \times 10^{-2}}\right)^2 \\ & =\frac{10^4}{4.9 \times 4.9}=\frac{100 \times 100}{24.01} \\ & =4.1649 \times 100 \\ & =416.49\end{aligned}$

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2006
The equilibrium constant for the reaction Zns+Sn2+aq⇌Zn2+aq+Sns is 1×1020 at 298 K. The magnitude of standard electrode potential of Sn/Sn2+ if EZn2*/Zn0=-0.76 V is _____ ×10-2 V. (Nearest integer)

Given : 2.303RTF=0.059 V

Options:

The correct answer is:
17

Solution:
Zns+Sn2+aq⇌Zn2+aq+Sns

Given: 2.303RTF = 0.059 V

Ecell0 = 2.303RT2F log Kc

⇒ Ecell0 = 0.0592 log (1020)

⇒ Ecell0 = 0.0592 × 20

⇒ Ecell0 = 0.59 V

Ecell0 = ESn2+/Sn0 - EZn2+/Zn0

⇒ Ecell0 = ESn2+/Sn0 - (-0.76)

⇒ ESn2+/Sn0 = -0.76 + 059 = -0.17

⇒ ESn/Sn2+0 = 0.17 = 17 × 10-2

So, answer is 17.

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2023 (29 Jan Shift 2)
The equilibrium constant, $K_C$ for $3 \mathrm{C}_2 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons \mathrm{C}_6 \mathrm{H}_6(\mathrm{~g})$ is $4 \mathrm{~L}^2 \mathrm{~mol}^{-2}$. If the equilil rium concentration of benzene is $0.5 \mathrm{~mol}^{-1}$ than what is the value of concentration of ethylene?

Options:
A $0.05 \mathrm{~mol} / \mathrm{L}$
B $0.5 \mathrm{~mol} / \mathrm{L}$
C $0.25 \mathrm{~mol} / \mathrm{L}$
D $0.025 \mathrm{~mol} / \mathrm{L}$

The correct answer is:
$0.5 \mathrm{~mol} / \mathrm{L}$

Solution:
$$
\begin{aligned}
& 3 \mathrm{C}_2 \mathrm{H}_2(\mathrm{~g}) \rightleftharpoons \rightleftharpoons \mathrm{C}_6 \mathrm{H}_6(\mathrm{~g}) \\
& \boldsymbol{K}_{\mathrm{C}}=\frac{\left[\mathrm{C}_6 \mathrm{H}_6\right]}{\left[\mathrm{C}_2 \mathrm{H}_2\right]^3}=\frac{0.5}{\left[\mathrm{C}_2 \mathrm{H}_2\right]^3} \\
& 4=\frac{0.5}{\left[\mathrm{C}_2 \mathrm{H}_2\right]^3} \\
& {\left[\mathrm{C}_2 \mathrm{H}_2\right] }=\sqrt[3]{\frac{0.5}{4}}=\frac{1}{2} \\
& {\left[\mathrm{C}_2 \mathrm{H}_2\right] }=0.5 \mathrm{~mol} \mathrm{~L}^{-1}
\end{aligned}
$$
Therefore, the concentration of ethylene is $0.5 \mathrm{~mol} / \mathrm{L}$.

Topic: Chemistry, Chemical Equilibrium, JEE Main
The equilibrium constant $\left(\mathrm{K}_{\mathrm{c}}\right)$ for the reaction $\mathrm{N}_2(\mathrm{g})+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{NO}(\mathrm{g})$ at temperature $\mathrm{T}$ is $4 \times 10^{-4}$. The value of $\mathrm{K}_{\mathrm{c}}$ for the reaction, $\mathrm{NO}(\mathrm{g}) \rightarrow{1 / 2} \mathrm{~N}_2(\mathrm{g})+{1 / 2} \mathrm{~O}_2(\mathrm{g})$ at the same temperature is :

Options:
A
$0.02$

B
$2.5 \times 10^2$

C
$4 \times 10^{-4}$

D
$50.0$

The correct answer is:
$50.0$

Solution:
$\begin{array}{ll}\mathrm{N}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{NO} & \mathrm{K}_{\mathrm{C}}=4 \times 10^{-4} \\ \mathrm{NO} \rightleftharpoons \frac{1}{2} \mathrm{~N}_2+\frac{1}{2} \mathrm{O}_2 & \mathrm{~K}_{\mathrm{C}}^1=\sqrt{\frac{1}{\mathrm{~K}_{\mathrm{C}}}} \\ \mathrm{K}_{\mathrm{C}}^1=\frac{1}{\sqrt{4 \times 10^{-4}}}=50 & \end{array}$

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2012 (Offline)
The equilibrium constant of the following isomerisation reaction at 400âK and 298âK are 2.07 and 3.42 respectively.

Which of the following is/are correct?
I. The reaction is exothermic
II. The reaction is endothermic
III. At 400âK 50% of cis-butene and 50% of trans-butene are present of equilibrium
IV. Both at 298âK and 400âK , k1 =kâ1

Options:
A I and IV
B II and IV
C I and III
D I only

The correct answer is:
I only

Solution:
As on increasing temperature Keq is increasing so reaction is exothermic.

Topic: Chemistry, Chemical Equilibrium, JEE Main
The equilibrium constants at 298 K for a reaction A+B⇌C+D is 100. If the initial concentration of all the four species were 1 M each, then equilibrium concentration of D (in mol L-1 ) will be

Options:
A 1.182
B 0.182
C 0.818
D 1.818

The correct answer is:
1.818

Solution:
A+B⇌C+D K=100
Q=1×11×1=1
Hence Q<K so reaction moves in forward reaction
(1+x)21-x2=100
1+x1-x=10
1+x=10-10x
x=911
So, =1+x=1+911=1.818 M

Topic: Chemistry, Chemical Equilibrium, JEE Main
The equilibrium constants $K_{P_1}$ and $K_{P_2}$ for the reactions $X \rightleftharpoons 2 Y$ and $Z \rightleftharpoons P+Q$, respectively are in the ratio of $1: 9$. If the degree of dissociation of $X$ and $Z$ be equal then the ratio of total pressure at these equilibria is

Options:
A
$1 : 36$

B
$1 : 1$

C
$1 : 3$

D
$1 : 9$

The correct answer is:
$1 : 36$

Solution:
$$
\begin{aligned}
& X \rightleftharpoons 2 Y \\
& 10 \\
& (1-x) \quad 2 x \\
& k_{p_1}=\frac{(2 x)^2}{(1-x)}\left(\frac{P_1}{1+x}\right)^1 \\
& Z \rightleftharpoons P+Q \\
& 100 \\
& (1-x) \quad x \quad x \\
& \mathrm{k}_{\mathrm{p}_2}=\frac{\mathrm{x}^2}{(1-\mathrm{x})}\left(\frac{\mathrm{P}_2}{1+\mathrm{x}}\right)^1 \\
& \frac{4 \times \mathrm{P}_1}{\mathrm{P}_2}=\frac{1}{9} \Rightarrow \frac{\mathrm{P}_1}{\mathrm{P}_2}=\frac{1}{36} \\
&
\end{aligned}
$$

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2008