Question

The standard enthalpy of the decomposition of $N _{2} O _{4}$ to $NO _{2}$ is $58.04\, kJ$ and standard entropy of this reaction is $176.7\, J / K$. Therefore the standard free energy change for this reaction at $25^{\circ} C$ is

• A. $-5.39\, kJ$
• B. $539\, kJ$
• C. $5.39\, kJ$
• D. $-539\, kJ$

Answer 1

Answer: (C)

Standard enthalpy of decompositions $(\Delta H)$

Standard entropy $(\Delta S)=176.7 J / K =176.7 \times 10^{-3} kJ / K $

and temperature $(T)=25^{\circ} C =298 K$

Change in free energy $(\Delta G)=\Delta H-T \Delta S$

$=58.04-298 \times\left(176.7 \times 10^{-3}\right)$

$=58.04-52.65=5.39\, kJ$