Question

Consider the following reaction at 35°C:
$\text{2NOCI}\left(g\right)\leftrightarrow 2NO\left(g\right)+C{l}_2\left(g\right)\Delta G^o=20kJ$
If 2.0 atm of NOClare reacted in a rigid container at 35°C, calculate the equilibrium partial pressure of NO .

Short answer

The equilibrium partial pressure of NO is 0.14atm.

Step-by-step solution

Introduction to the Concept

The Gibbs-Helmholtz isotherm links the Gibbs free energy of a reaction at equilibrium to the equilibrium constant (K):

$\Delta G_R°=-RT\ln K-----\left(1\right)$

Where $R=8.314J/mol-K$ and T is the temperature.

Determination of the equilibrium constant of K

From the given reaction,

$2NOCI\left(g\right)\leftrightarrow 2NO\left(g\right)+C{l}_2\left(g\right)\Delta G^o=20kJ$

The free energy change of $NOCI=\frac{20\text{kJ}}{2\text{mole}}$

$=10KJ/mol$

From equation (1),

$\Delta G_R°=-RT\ln K$

$K=e^{-\frac{\Delta G_R°}{RT}}$

$=e^{-\frac{10KJ/mol}{0.08314KJ/mol-k\times (35+273)K}}$

$=2.01\times {10}^{-2}$

Determination of the equilibrium partial pressure of  NO

$2NOCI\left(g\right)\leftrightarrow 2NO\left(g\right)+C{l}_2\left(g\right)$

$K_{eq}=\frac{P_{NO}^2\times P_{C{l}_2}}{P_{NOCl}^2}$

$=\frac{{\left(2x\right)}^2\times \left(x\right)}{{\left(2-2x\right)}^2}$

$2.01\times {10}^{-2}=\frac{4x^3}{{\left(2-2x\right)}^2}$

$x=0.136\text{atm}$

$P_{NO}\simeq 0.14\text{atm}$

Therefore the equilibrium partial pressure is 0.14atm .