Question

From application of thermodynamics on chemical reaction,
we get $\Delta G=\Delta G^0+RT$ ln Q
Also $\Delta G=\Delta H-T\Delta S$.
If $\Delta G=0$, reaction is at equilibrium.
If $\Delta G>0$ reaction is non-spontaneous under given condition. $If\Delta G<0$, reaction is spontaneous under given condition.

Consider the reaction given below.
$A\left(s\right)\rightleftharpoons 2B\left(g\right)\Delta H^0=160$ KJ/mol.
$\Delta S^0=400$ J/mol-K at $400$ K
Which of the following is correct at $400$ K?

• A. On adding more $A\left(s\right)$, more $B\left(g\right)$ is produced, when $A\left(s\right)$ and $B\left(g\right)$ were in equilibrium
• B. The equilibrium constant at $400$ K can't be found
• C. The reaction is at equilibrium at $400$ K and standard condition
• D. The $\Delta G$ of the reaction is greater than zero, at $400$ K and standard condition

Answer 1

The correct option is C The reaction is at equilibrium at $400$ K and standard condition

Th relationship between the standard free energy change, standard enthalpy change and entropy change is $\Delta G^o=\Delta H^o-T\Delta S^o$.
$\Delta H^o=160kJ/mol=160000J/mol$.
Substitute values in the above expression.
$\Delta G^o=160000J/mol-400K\times 400J/mol-K=0J/mol$.
Under standard conditions, $K_p=1$, so $ln{K}_p=0$.
Thus $\Delta G=\Delta G^0+RTln{K}_p=0+RT\left(0\right)=0$.
Hence, the system is at equilibrium when a temperature is 400 K and standard conditions are used.