Thermodynamics and Thermochemistry - Result Question 70

####70. Show that the reaction, $CO(g)+\frac{1}{2} O _2(g) \longrightarrow CO _2(g)$ at $300 K$, is spontaneous and exothermic, when the standard entropy change is $-0.094 kJ mol^{-1} K^{-1}$. The standard Gibbs’ free energies of formation for $CO _2$ and $CO$ are -394.4 and $-137.2 kJ mol^{-1}$, respectively.

(2000, 3M)

Show Answer

Solution:

  1. $\Delta _r G^{\circ}=\Delta _f G^{\circ}$ (products) $-\Delta _f G^{\circ}$ (reactants)

$$ =-394.4-(-137.2)=-257.2 kJ<0 $$

The above negative value of $\Delta G$ indicates that the process is spontaneous.

Also, $\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}$

$$ \begin{aligned} \Rightarrow \quad \Delta H^{\circ} & =\Delta G^{\circ}+T \Delta S^{\circ} \ & =-257.2+300(-0.094) \ & =-285.4 kJ<0 \end{aligned} $$



NCERT Chapter Video Solution

Dual Pane