(A) For a reversible reaction $A \rightleftharpoons B$,the equilibrium constant $K$ is related to the enthalpy change $\Delta H$ by the van't Hoff equ

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(A) For a reversible reaction $A \rightleftharpoons B$,the equilibrium constant $K$ is related to the enthalpy change $\Delta H$ by the van't Hoff equation: $\ln K = -\frac{\Delta H}{RT}$.Given $\Delta H = -41.5 \ kJ \ mol^{-1}$,$R = 8.314 \times 10^{-3} \ kJ \ mol^{-1} \ K^{-1}$,and $T = 500 \ K$.Substituting the values: $\ln K = -\frac{-41.5}{8.314 \times 10^{-3} \times 500}$.$\ln K = \frac{41.5}{4.157} \approx 9.983$.Therefore,$K = e^{9.983} \approx e^{10}$.

Class 11 Chemistry 6-1.Equilibrium (Chemical Equilibrium)Kp and Kc Relationship

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For a reversible reaction $A \rightleftharpoons B$,the pre-exponential factor is the same for both the forward and backward reactions and has a value of $20 \ s^{-1}$. If the enthalpy change for the forward reaction is $-41.5 \ kJ \ mol^{-1}$,the value of the equilibrium constant at $500 \ K$ is:

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$e^{10}$
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$e^9$
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$e^8$
D
$e^7$

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For a reversible reaction $A \rightleftharpoons B$,the pre-exponential factor is the same for both the forward and backward reactions and has a value of $20 \ s^{-1}$. If the enthalpy change for the forward reaction is $-41.5 \ kJ \ mol^{-1}$,the value of the equilibrium constant at $500 \ K$ is:

Similar Questions

When $2 \, mol$ of $PCl_5$ is heated in a closed vessel of $2 \, L$ capacity,$40\%$ of $PCl_5$ dissociates into $PCl_3$ and $Cl_2$ at equilibrium. The value of the equilibrium constant $K_c$ will be:

For the equilibrium,$2 \ NOCl \ (g) \rightleftharpoons 2 \ NO \ (g) + Cl_{2} \ (g)$,the value of the equilibrium constant,$K_{c}$ is $3.75 \times 10^{-6}$ at $1069 \ K$. Calculate the $K_{p}$ for the reaction at this temperature?

The equilibrium concentrations of $X$,$Y$,and $YX_2$ are $4 \ mol/L$,$2 \ mol/L$,and $2 \ mol/L$ respectively for the equilibrium $2X + Y \rightleftharpoons YX_2$. The value of $K_c$ is:

One mole of $SO_3$ was placed in a $1 \ L$ reaction vessel at a certain temperature. The following equilibrium was established: $2SO_3 \rightleftharpoons 2SO_2 + O_2$. At equilibrium,$0.6 \ moles$ of $SO_2$ were formed. The equilibrium constant $(K_c)$ of the reaction will be:

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