At what temperature (T K) will the reaction | vedclass

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

Question

(A) For a reaction to be in equilibrium,the change in Gibbs free energy $(\Delta G)$ must be zero.Using the relation $\Delta G = \Delta H - T \Delta S

Vedclass · Accepted Answer

$462.12$

Vedclass · Answer

(A) For a reaction to be in equilibrium,the change in Gibbs free energy $(\Delta G)$ must be zero.Using the relation $\Delta G = \Delta H - T \Delta S$,at equilibrium $\Delta G = 0$,so $\Delta H = T \Delta S$.Given $\Delta H = 30.5 \ kJ \ mol^{-1}$ and $\Delta S = 0.066 \ kJ \ K^{-1} \ mol^{-1}$.Substituting the values: $30.5 = T 	imes 0.066$.$T = \frac{30.5}{0.066} \approx 462.12 \ K$.

At what temperature $(T \ K)$ will the reaction be in equilibrium (in $K$)? $Ag_2O_{(s)} \rightarrow 2Ag_{(s)} + \frac{1}{2} O_{2(g)}$,given $\Delta H = 30.5 \ kJ \ mol^{-1}$ and $\Delta S = 0.066 \ kJ \ K^{-1} \ mol^{-1}$.

Similar Questions

Which of the following relations is correct?

$A$ reaction has both $\Delta H$ and $\Delta S$ negative. The rate of reaction

State the criteria for a spontaneous process in terms of $\Delta G$ at constant pressure and temperature.

For a spontaneous change,free energy change $\Delta G$ is

Calculate the work done in the following reaction at $27^{\circ} C$.
$4 SO_{2(g)} + 2 O_{2(g)} \rightarrow 4 SO_{3(g)}$
$(R = 8.314 \ J \ K^{-1} \ mol^{-1})$ (in $J$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module