Subject: Chemistry — Chemical Thermodynamics
Question: The surface of copper gets tarnished by the formation of
copper oxide. $N_{2}$ gas was passed to prevent the oxide formation
during heating of copper at 1250 K. However, the $N_{2}$ gas contains 1
mole % of water vapour as impurity. The water vapour oxidises copper as
per the reaction given below:
$$2Cu(s) + H_{2}O(g) \rightarrow Cu_{2}O(s) + H_{2}(g)$$
$pH_{2}$ is the minimum partial pressure of $H_{2}$ (in bar) needed to
prevent the oxidation at 1250 K. The value of
$\ln\left( {p_{H}}_{2} \right)$ is \_\_\_\_\_\_\_\_\_.
(Given: total pressure = 1 bar, R (universal gas constant)
$= 8\ J\ K^{- 1}\ mol^{- 1},\ln(10) = 2.3,Cu(s)$ and $Cu_{2}O(s)$ are
mutually immiscible.
At
$1250K:2Cu(s) + \frac{1}{2}O_{2}(g) \rightarrow Cu_{2}O(s);\mathrm{\Delta}G{^\circ} = - 78,000\ J\ mol^{- 1}$
$H_{2}(g) + \frac{1}{2}O_{2}(g)\backslash righattow\ H_{2}O(g);\mathrm{\Delta}G{^\circ} = - 1,78,000\ J\ mol^{- 1};G$
is the Gibbs energy)
Answer: --14.6
Solution: (i)
$2Cu(s) + \frac{1}{2}O_{2}(g) \rightarrow Cu_{2}O(s):\mathrm{\Delta}G{^\circ} = - 78000\ J/mol$
\(ii\)
$H_{2(g)} + \frac{1}{2}O_{2}(g) \rightarrow H_{2}O(g),\mathrm{\Delta}G{^\circ} = - 178000\ J/mol$
\(i\) -- (ii) then
$$2Cu(s) + H_{2}O(g) \rightarrow Cu_{2}O(s) + H_{2}(g)$$
$$\mathrm{\Delta}G{^\circ} = - 78000 + 178000 = 100000\ J/mol$$
Now for the above reaction
$$\mathrm{\Delta}G = \mathrm{\Delta}G{^\circ} + RT\ln\left( \frac{{P_{H}}_{2}}{P_{H_{2}O}} \right)$$
To prevent the above reaction :
$$\mathrm{\Delta}G \geq 0$$
$$\mathrm{\Delta}G{^\circ} + RT\ln\left( \frac{P_{H_{2}}}{P_{H_{2}O}} \right) \geq 0$$
$$10^{4}\left( \ln P_{H_{2}} - \ln P_{H_{2}O} \right) \geq - 10^{5}$$
$$\ln P_{H_{2}} \geq - 10 + \ln P_{H_{2}O}$$
$$\geq - 10 + 2.3\log(0.01)$$
$$\ln P_{H_{2}} \geq - 10 + 4.6$$
$$\ln P_{H_{2}} \geq - 14.6$$
∴Minimum $\ln P_{H_{2}} = - 14.6$