Question

Assume $△H^{o}and△S^o$ to be independent of temperature, at what temperature will the reaction given below becomes spontaneous?
$N_2\left(g\right)+O_2\left(g\right)\to 2NO\left(g\right)$
$S_{N_2\left(g\right)}^o=191.4J{K}^{-1}mo{l}^{-1}$
$S_{O_2\left(g\right)}^o=204.9J{K}^{-1}mo{l}^{-1}$
$S_{NO\left(g\right)}^o=210.5J{K}^{-1}mo{l}^{-1}$
$△H^o=180.8kJmo{l}^{-1}$

• A. 7320 K
• B. 3650 K
• C. 6595 K
• D. 8500 K

Answer 1

The correct option is A 7320 K

Given, $N_2\left(g\right)+O_2\left(g\right)\to 2NO\left(g\right)$
$S_{N_2\left(g\right)}^o=191.4J{K}^{-1}mo{l}^{-1}$
$S_{O_2\left(g\right)}^o=204.9J{K}^{-1}mo{l}^{-1}$
$S_{NO\left(g\right)}^o=210.5J{K}^{-1}mo{l}^{-1}$
$△H^o=180.8kJmo{l}^{-1}$
The entropy of the reaction can be calculated as
$△S^o=\left(2S_{NO}^o\right)-(S_{N_2}^o+S_{O_2}^o)$
putting the values, we get
$=(2\times 210.5)-(191.4+204.9)$
$=24.7J{K}^{-1}mo{l}^{-1}$
Also, we know gibb's free energy relation,
$△G^o=△H^o-T△S^o$
Putting the values, we get;
$△G^o=180.8-(T\times 24.7\times {10}^{-3})$
Reaction becomes spontaneous, when
$△G^o<0$
$180.8-T\times 24.7\times {10}^{-3}<0$
which means,
$T>\frac{180.8}{24.7}\times {10}^3=7320K$