JEE Advance - Chemistry (2018 - Paper 2 Offline - No. 7)

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 vapor as impurity. The water vapor oxidises copper as per the reaction given below : $$2Cu\left( s \right) + {H_2}O\left( g \right) \to C{u_2}O\left( s \right) + {H_2}\left( g \right)$$

$${P_{H2}}$$ is the minimum partial pressure of $${H_2}$$ (in bar) needed to prevent the oxidation at $$1250$$ $$K.$$ The value of $$\ln \left( {{P_{H2}}} \right)$$ is ________.

Given: total pressure $$=1$$ bar, $$R$$ (universal gas constant ) $$=$$ $$8J{K^{ - 1}}\,\,mo{l^{ - 1}},$$ $$\ln \left( {10} \right) = 2.3.\,$$ $$Cu(s)$$ and $$C{u_2}O\left( s \right)$$ are naturally immiscible.

At $$1250$$ $$K:2Cu(s)$$ $$ + {\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}{O_2}\left( g \right) \to C{u_2}O\left( s \right);$$ $$\Delta {G^ \circ } = - 78,000J\,mo{l^{ - 1}}$$

$${H_2}\left( g \right) + {\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}{O_2}\left( g \right) \to {H_2}O\left( g \right);$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$\Delta {G^ \circ } = - 1,78,000J\,mo{l^{ - 1}};$$ ($$G$$ is the Gibbs energy)
Answer
-14.6

Explanation

$$2Cu(s) + {1 \over 2}{O_2}(g) \to C{u_2}O(s);\Delta {G^o} = - 78,000$$ J mol$$-$$1 ...... (1)

$${H_2}(g) + {1 \over 2}{O_2}(g) \to {H_2}O(g);\Delta {G^o} = - 1,78,000$$ J mol$$-$$1 ..... (2)

Subtracting Eq. (1) $$-$$ Eq. (2), we get

$$2Cu(s) + {H_2}O(g) \to C{u_2}O(s) + {H_2}(g);\Delta {G^o} = + 100000$$ J mol$$-$$1 ..... (3)

Now, $$\Delta G = \Delta {G^o} + RT\ln \left( {{{{p_{{H_2}}}} \over {{p_{{H_2}O}}}}} \right)$$

For the reaction (3) to not occur,

$$\Delta G > 0$$ or $$\Delta {G^o} + RT\ln \left( {{{{p_{{H_2}}}} \over {{p_{{H_2}O}}}}} \right) > 0$$

$$ \Rightarrow 100000 + 8 \times 1250\ln \left( {{{{p_{{H_2}}}} \over {{p_{{H_2}O}}}}} \right) > 0$$

$$ \Rightarrow 100000 + 8 \times 1250\ln \left( {{{{p_{{H_2}}}} \over {{p_{{H_2}O}}}}} \right) > {{ - 100000} \over {8 \times 1250}}$$

$$ \Rightarrow \ln \left( {{{{p_{{H_2}}}} \over {{p_{{H_2}O}}}}} \right) > - 10$$

$$\ln {p_{{H_2}}} > - 10 + \ln {p_{{H_2}O}}$$

Now, $${p_{{H_2}O}} = {x_{{H_2}O}} \times {p_T} = 0.01 \times 1 = {10^{ - 2}}$$

So, $$\ln {p_{{H_2}}} > - 10 - 2\ln 10 \Rightarrow {p_{{H_2}}} > - 14.6$$ bar

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