(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 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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A
$e^{10}$
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$e^9$
C
$e^8$
D
$e^7$

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1 Equilibrium: Chemical Equilibrium

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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:

Similar Questions

For the gaseous reaction $aA + bB \rightleftharpoons cC + dD$,the relation between $K_p$ and $K_c$ is:

The value of $K_{p}$ for the reaction,$CO_{2(g)} + C_{(s)} \rightleftharpoons 2CO_{(g)}$ is $3.0$ at $1000 \ K$. If initially $P_{CO_{2}} = 0.48 \ bar$ and $P_{CO} = 0 \ bar$ and pure graphite is present,calculate the equilibrium partial pressures of $CO$ and $CO_{2}$.

For the reaction $A + B \rightleftharpoons C + D$ at $298 \ K$,the equilibrium constant is $10.0$. If the initial concentration of all four species is $1 \ M$,what will be the equilibrium concentration of $D$ (in $mol \ L^{-1}$)?

If for a reaction $\Delta G^{o} > 0$,then:

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