Question

For the reaction
$N_2\left(g\right)+3H_2\left(g\right)\to 2N{H}_3\left(g\right)$
$△H=-95.4kJand△S=-198.3J{K}^{-1}$
Calculate the temperature at which Gibb's energy change $(△G$) is equal to zero.

• A. 495 K
• B. 481 K
• C. 430 K
• D. 450 K

Answer 1

The correct option is B 481 K

We know, $△G=△H-T△S$
So, for $△G=0$
we have $△H-T△S=0$
$or△H=T△S$
$T=\frac{△H}{△S}$
$=\frac{-95.4\times 1000J}{-198.3J{K}^{-1}}$
$=481K$

At this temperature, the reaction would be in equilibrium and with the increase in temperature the opposing factor $T△S$ would become more and hence, $△G$ would become positive and the reaction would become non-spontaneous. The reaction would be spontaneous at the temperature below 481 K.