(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

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(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 in entropy $(\Delta S)$:$\Delta S = S_{AB_3} - (\frac{1}{2} S_{A_2} + \frac{3}{2} S_{B_2})$$\Delta S = 50 - (\frac{1}{2} \times 60 + \frac{3}{2} \times 40)$$\Delta S = 50 - (30 + 60) = 50 - 90 = -40 \, J K^{-1} mol^{-1}$.At equilibrium,$\Delta G = 0$,so $\Delta H = T \Delta S$.$T = \frac{\Delta H}{\Delta S} = \frac{-20000 \, J}{-40 \, J K^{-1}} = 500 \, K$.

Class 11 Chemistry 1 Equilibrium: Chemical Equilibrium Standard free energy

Difficult

For a hypothetical reversible reaction $\frac{1}{2} A_{2(g)} + \frac{3}{2} B_{2(g)} \rightarrow AB_{3(g)}$; $\Delta H = -20 \, kJ$. If the standard entropies of $A_2, B_2$ and $AB_3$ are $60, 40$ and $50 \, J K^{-1} mol^{-1}$ respectively,at what temperature $(K)$ will the reaction be in equilibrium?

A
$400$
B
$500$
C
$250$
D
$200$

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Class 11

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Chapter

1 Equilibrium: Chemical Equilibrium

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Standard free energy

The equilibrium constant $(K)$ of a reaction may be written as :

EasyWBJEE 2009

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For the reaction at $25\,^oC$,${N_2O_4}_{(g)} \rightleftharpoons 2NO_{2_{(g)}}$,if $\Delta G_f^o$ for $N_2O_4$ and $NO_2$ are $23.49 \, KCal$ and $12.39 \, KCal$ respectively,then $K_p$ for the reaction is: (in $, atm$)

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Explain chemical equilibrium and free Gibbs energy change by $\Delta G$.

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The value of $\log _{10} K$ for a reaction $A \rightleftharpoons B$ is
(Given : $\Delta _{r} H_{298 K}^{\circ} = -54.07 \ kJ \ mol^{-1}$,$\Delta _{r} S_{298 K}^{\circ} = 10 \ J \ K^{-1} \ mol^{-1}$ and $R = 8.314 \ J \ K^{-1} \ mol^{-1}$; $2.303 \times 8.314 \times 298 = 5705$)

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The $\Delta G^o$ for the reaction $X + Y \rightleftharpoons Z$ is $-4.606 \ kcal$. The value of the equilibrium constant of the reaction at $227 \ ^oC$ is $(R = 2.0 \ cal \ mol^{-1} K^{-1})$.

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The equilibrium constant $(K)$ of a reaction may be written as :

For the reaction at $25\,^oC$,${N_2O_4}_{(g)} \rightleftharpoons 2NO_{2_{(g)}}$,if $\Delta G_f^o$ for $N_2O_4$ and $NO_2$ are $23.49 \, KCal$ and $12.39 \, KCal$ respectively,then $K_p$ for the reaction is: (in $, atm$)

Explain chemical equilibrium and free Gibbs energy change by $\Delta G$.

The value of $\log _{10} K$ for a reaction $A \rightleftharpoons B$ is(Given : $\Delta _{r} H_{298 K}^{\circ} = -54.07 \ kJ \ mol^{-1}$,$\Delta _{r} S_{298 K}^{\circ} = 10 \ J \ K^{-1} \ mol^{-1}$ and $R = 8.314 \ J \ K^{-1} \ mol^{-1}$; $2.303 \times 8.314 \times 298 = 5705$)

The $\Delta G^o$ for the reaction $X + Y \rightleftharpoons Z$ is $-4.606 \ kcal$. The value of the equilibrium constant of the reaction at $227 \ ^oC$ is $(R = 2.0 \ cal \ mol^{-1} K^{-1})$.

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The value of $\log _{10} K$ for a reaction $A \rightleftharpoons B$ is
(Given : $\Delta _{r} H_{298 K}^{\circ} = -54.07 \ kJ \ mol^{-1}$,$\Delta _{r} S_{298 K}^{\circ} = 10 \ J \ K^{-1} \ mol^{-1}$ and $R = 8.314 \ J \ K^{-1} \ mol^{-1}$; $2.303 \times 8.314 \times 298 = 5705$)