Question

Standard Free Energy and Equilibrium constant: The change in free energy for a reaction taking place between gaseous reactants and products represented by general equation.
$\Delta G=\Delta G^{\circ }+RTln{Q}_P$ the condition for a system to be at equilibrium is that
$\Delta G=0\text{and}{Q}_p=K_P$
Thus at equilibrium
$\Delta G^{\circ }=-RTln{K}_P$
Note: In the reaction, where all gaseous reactants and products; $K\text{represents}{K}_P$, we may conclude that for standard reactions, i.e., at $1\text{M}\text{or}1\text{atm}$
When $\Delta G^{\circ }=-ve\text{or}K>1:\text{forward reaction is feasible}$
$\Delta G^{\circ }=+ve\text{or}K<1:\text{reverse reaction is feasible}$
$\Delta G^{\circ }=0\text{or}K=1:\text{reaction is at equilibrium (very rare)}$
For the equilibrium $NiO\left(s\right)+CO\left(g\right)\rightleftharpoons Ni\left(s\right)+C{O}_2\left(g\right),\Delta G^{\circ }\left({\text{J Mol}}^{-1}\right)=-20700-11.97T.$. Calculate the temperature at which the product gases at equilibrium at $1\text{atm}$ will conatin $400\text{ppm}$ of carbon monoxide.

• A. $390\text{K}$
• B. $400\text{K}$
• C. $275\text{K}$
• D. $200\text{K}$

Answer 1

The correct option is A $390\text{K}$

$\text{For the given reaction}{P}_{CO}<<P_{C{O}_2}$
$K_P=\frac{1}{400\times {10}^{-6}}=2500$
$\Delta G^{\circ }=-RTln{K}_P$
$\Rightarrow ln{K}_P=-\frac{\Delta G^{\circ }}{RT}=\frac{20700+11.97T}{RT}$
$\Rightarrow \left(ln2500\right)\times \left(8.314T\right)=20700+11.97T$
$\Rightarrow 65T=20700+11.97T$
$\Rightarrow 53.03T=20700$
$\Rightarrow T=\frac{20700}{53.03}=390\text{K}$