Question

The surface of copper gets tarnished by the formation of copper oxide. ${{N}_{2}}$ 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
\[2Cu\left( s \right)+{{H}_{2}}O\left( g \right)\to C{{u}_{2}}O\left( s \right)+{{H}_{2}}\left( g \right)\]
${{\rho }_{{{H}_{2}}}}$ ​​ is the minimum partial pressure of ${{H}_{2}}$ (in bar) needed to prevent the oxidation at 1250 K. The value of $\ln ({{\rho }_{{{H}_{2}}}})$ is _________.
Given:
Total pressure =1 bar
R (universal gas constant) =8$~J{{K}^{-1}}mo{{l}^{-1}}$
ln(10) = 2.3
Cu(s) and $C{{u}_{2}}O$ (s) are mutually immiscible
At 1250 K:
\[\begin{align}
  & 2Cu\left( s \right)+\dfrac{1}{2}{{O}_{2}}\left( g \right)\to C{{u}_{2}}O\left( s \right);~\Delta {{G}^{\circ }}=-78,000~J~mo{{l}^{-1}}~ \\
 & {{H}_{2}}\left( g \right)+\dfrac{1}{2}{{O}_{2}}\left( g \right)\to {{H}_{2}}O\left( g \right);~\Delta {{G}^{\circ }}=-1,78,000~J~mo{{l}^{-1}} \\
\end{align}\]
 [Upto one decimal point and take modulus of the answer]

Answer 1

Hint: By using the below formula we can calculate $\ln ({{\rho }_{{{H}_{2}}}})$ .
 \[\Delta G=\Delta {{G}^{\circ }}+RT~ln~Q\]
$\Delta G$ = Gibbs free energy
$\Delta {{G}^{o}}$ = free energy releases during the chemical reaction
R = Rydberg constant
T = Temperature
Q = $\ln \left( \dfrac{{{\rho }_{{{H}_{2}}}}}{{{\rho }_{{{H}_{2}}O}}} \right)$
Here ${{\rho }_{{{H}_{2}}}}$ = minimum partial pressure of hydrogen
${{\rho }_{{{H}_{2}}}}_{O}$ = partial pressure of water.

Complete answer:
- In the question they have given the free energies of the different reactions and they are as follows.
\[\begin{align}
  & 2Cu\left( s \right)+\dfrac{1}{2}{{O}_{2}}\left( g \right)\to C{{u}_{2}}O\left( s \right);~\Delta {{G}^{\circ }}=-78,~KJ~~ \\
 & {{H}_{2}}\left( g \right)+\dfrac{1}{2}{{O}_{2}}\left( g \right)\to {{H}_{2}}O\left( g \right);~\Delta {{G}^{\circ }}=-1,78KJ~mol \\
\end{align}\]
From the above equations we can write the following chemical reaction
\[2Cu(s)+{{H}_{2}}O(g)\to C{{u}_{2}}O+{{H}_{2}}(g)~,\Delta {{G}^{\circ }}=+100~kJ\]
Substitute the above Gibbs energy value in the below formula to get the value of $\ln ({{\rho }_{{{H}_{2}}}})$.
\[\Delta G=\Delta {{G}^{\circ }}+RT~ln~Q\]
$\Delta G$ = Gibbs free energy
$\Delta {{G}^{o}}$ = free energy releases during the chemical reaction
R = Rydberg constant
T = Temperature
Q = $\ln \left( \dfrac{{{\rho }_{{{H}_{2}}}}}{{{\rho }_{{{H}_{2}}O}}} \right)$
\[\begin{align}
  & \Delta G=\Delta {{G}^{\circ }}+RT~ln~Q \\
 & 0=100+0.008\times 1250\times \ln \left( \dfrac{{{\rho }_{{{H}_{2}}}}}{{{\rho }_{{{H}_{2}}}}_{O}} \right) \\
 & \ln {{\rho }_{{{H}_{2}}}}=-14.6 \\
\end{align}\]
Here ${{\rho }_{{{H}_{2}}}}$ = minimum partial pressure of hydrogen
- Therefore the value of $\ln ({{\rho }_{{{H}_{2}}}})$ is -14.6.

Note:
We have to consider the overall reaction of copper with water to form cupric oxide and hydrogen to calculate the Gibbs free energy of the chemical reaction. We are not supposed to consider individual Gibbs free energy while calculating the partial pressure of hydrogen.