The standard enthalpy of the decomposition of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The standard free energy change is calculated using the Gibbs-Helmholtz equation: $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$. Given: $\De

Vedclass · Accepted Answer

$5.39$

Vedclass · Answer

(D) The standard free energy change is calculated using the Gibbs-Helmholtz equation: $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$. Given: $\Delta H^\circ = 58.04 \, kJ = 58040 \, J$,$\Delta S^\circ = 176.7 \, J/K$,and $T = 25 + 273 = 298 \, K$. Substituting the values: $\Delta G^\circ = 58040 \, J - (298 \, K 	imes 176.7 \, J/K)$. $\Delta G^\circ = 58040 \, J - 52656.6 \, J = 5383.4 \, J$. Converting to $kJ$: $\Delta G^\circ = 5.3834 \, kJ \approx 5.39 \, kJ$.

The standard enthalpy of the decomposition of $N_2O_4$ to $NO_2$ is $58.04 \, kJ$ and standard entropy of this reaction is $176.7 \, J/K$. The standard free energy change for this reaction at $25 \, ^oC$ is $..... \, kJ$

Similar Questions

$A$ process is taking place at constant temperature and pressure. Then

For a given chemical reaction $A \rightarrow B$ at $300 \ K$,the free energy change is $-49.4 \ kJ \ mol^{-1}$ and the enthalpy of reaction is $51.4 \ kJ \ mol^{-1}$. The entropy change of the reaction is $..... \ J \ K^{-1} \ mol^{-1}$.

For a reaction at $25 \, ^oC$,the enthalpy change and entropy change are $-11.7 \times 10^3 \, J \, mol^{-1}$ and $-105 \, J \, mol^{-1} \, K^{-1}$ respectively. What is the Gibbs free energy in $kJ$?

At a temperature of $25^\circ C$,if $\Delta H < T \Delta S$,then $\Delta G$ = ?

The value of ${\left[ {\frac{dG}{dT}} \right]_P}$ is ....

Vedclass Test Series

Exam Paper Generator

Online Exam Module