For a hypothetical reversible reaction frac{1 | vedclass

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

Question

(B) The reaction is $\frac{1}{2} A_{2(g)} + \frac{3}{2} B_{2(g)} \rightarrow AB_{3(g)}$ with $\Delta H = -20 \, kJ = -20000 \, J$.Calculate the change

Vedclass · Accepted Answer

$500$

Vedclass · Answer

(B) The reaction is $\frac{1}{2} A_{2(g)} + \frac{3}{2} B_{2(g)} ightarrow AB_{3(g)}$ with $\Delta H = -20 \, kJ = -20000 \, J$.Calculate the change in entropy $(\Delta S)$:$\Delta S = S_{AB_3} - (\frac{1}{2} S_{A_2} + \frac{3}{2} S_{B_2})$$\Delta S = 50 - (\frac{1}{2} 	imes 60 + \frac{3}{2} 	imes 40)$$\Delta S = 50 - (30 + 60) = 50 - 90 = -40 \, J K^{-1} mol^{-1}$.At equilibrium,$\Delta G = 0$,so $\Delta H = T \Delta S$.$T = \frac{\Delta H}{\Delta S} = \frac{-20000 \, J}{-40 \, J K^{-1}} = 500 \, K$.

Similar Questions

Calculate the standard Gibbs free energy change $\Delta G^o$ at $298 \ K$ for the conversion of oxygen to ozone,given by the reaction: $\frac{3}{2} O_{2(g)} \rightleftharpoons O_{3(g)}$. The equilibrium constant $K_p$ for this conversion is $3 \times 10^{-29}$.

$A$ reaction attains equilibrium when the free energy change accompanying it is

Calculate $\Delta G^\circ$ for the reaction, $CH_4(g) + H_2(g) \rightarrow C_2H_6(g)$ at $298 \text{ K}$, given $K_p = 2 \times 10^{17}$ and $R = 8.314 \text{ J K}^{-1} \text{mol}^{-1}$.

In the equilibrium state,the value of $\Delta G$ is:

The correct relationship between the equilibrium constant $(K)$ and the standard Gibbs free energy change $(\Delta G^o)$ for a reaction is .......

Vedclass Test Series

Exam Paper Generator

Online Exam Module