JEE Advance - Chemistry (2018 - Paper 2 Offline - No. 7)
$${P_{H2}}$$ 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)
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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