Question

The free energy of formation of NO is 78 kJ mol^-1 at the temperature of an automobile engine (1000 K). What is the equilibrium constant for this reaction at 1000 K?

$\frac{1}{2} \mathrm{~N}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{NO}(\mathrm{g})$

(Anti log 0.9263 = 8.459)

• A. $8.4\times 10^{- 5}$
• B. $7.1\times 10^{- 9}$
• C. $4.2\times 10^{- 10}$
• D. $1.7\times 10^{- 19}$

Answer 1

Answer: (A)

$K_{e q}=antilog\left(\frac{- \Delta G^{o}}{2.303 R T}\right)$

$\Delta G^{o}=78 \, kJ/mol=78000 \, J/mol$

$R=8.314 \, JK^{- 1}mol^{- 1},T=1000 \, K$

Putting these values in eqn. (i) we get,

$K_{e q}=antilog\left[\frac{- 78000}{2.303 \times 8.314 \times 1000}\right]$

$K_{e q}=antilog\left(\right.-4.0737\left.\right)$ or $K_{e q}=8.4\times 10^{- 5}$