QJEEAdv18Ch4c9d: JEE Advanced 2018 Chemistry - Class 11 JEEAdv18Ch4c9d Solution: The surface of copper gets tarnished by the formation of copper …

Question

Verified Solution • 2025–26 Session moderate

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)

+ 1 / 2 O_{2} (g) \to C u_{2} O (s);

\Delta {G^ \circ } = - 78,000J\,mo{l^{ - 1}}

H_{2} (g) + 1 / 2 O_{2} (g) \to H_{2} O (g);

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,

\Delta {G^ \circ } = - 1,78,000J\,mo{l^{ - 1}};

(

G

is the Gibbs energy)

Accepted Answer

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