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 $1000 \,K$?
$\frac{1}{2} N_{2}(g)+\frac{1}{2} O_{2}(g) N O(g)$

• 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)

Free energy and equilibrium constant are related as,
$\Delta G=-2.303 R T \log K$
$78 \times 10^{3}=-2.303 \times 8.314 \times 1000 \times \log K$
$\log K=-\frac{78}{2.303 \times 8.314}$
$=-4.073$
$K =$ antilog $(-4.0 .73)$
$=8.43 \times 10^{-5}$