The reaction,CaCO_{3(s)} rightleftharpoons Ca | vedclass

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(A) The standard Gibbs free energy change for the reaction is calculated as: $\Delta G^{\circ} = \Delta G_{f}^{\circ}(CaO) + \Delta G_{f}^{\circ}(CO_{

Vedclass · Accepted Answer

$1.13 	imes 10^{-23}$

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(A) The standard Gibbs free energy change for the reaction is calculated as: $\Delta G^{\circ} = \Delta G_{f}^{\circ}(CaO) + \Delta G_{f}^{\circ}(CO_{2}) - \Delta G_{f}^{\circ}(CaCO_{3})$.Substituting the given values: $\Delta G^{\circ} = (-603.501) + (-394.389) - (-1128.79) = +130.9 \, kJ \, mol^{-1} = 130900 \, J \, mol^{-1}$.Using the relation $\Delta G^{\circ} = -RT \ln K_{p}$,we have $\ln K_{p} = -\frac{\Delta G^{\circ}}{RT}$.$\ln K_{p} = -\frac{130900}{8.314 	imes 298} \approx -52.834$.Converting to base $10$ logarithm: $\log_{10} K_{p} = \frac{-52.834}{2.303} \approx -22.941$.Thus,$K_{p} = 10^{-22.941} \approx 1.13 	imes 10^{-23}$.Since $K_{p} = P_{CO_{2}}$,the partial pressure of $CO_{2(g)}$ is $1.13 	imes 10^{-23} \, atm$.

Similar Questions

The equilibrium constant of the following given reaction is $K_p = 6.022 \times 10^{-5}$ at $298 \ K$ temperature. $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$. Calculate the value of $\Delta_r G^o$.

At $300 \ K$,$\Delta_{r} G^{\circ}$ for the reaction $A_{2(g)} \rightleftharpoons B_{2(g)}$ is $-11.5 \ kJ \ mol^{-1}$. The equilibrium constant at $300 \ K$ is approximately $(R=8.314 \ J \ mol^{-1} \ K^{-1})$.

Find the value of the equilibrium constant $(K)$ of a reaction at $300 \ K$, when standard Gibbs free energy change is $-25 \ kJ \ mol^{-1}$? (Consider $R = 8.33 \ J \ mol^{-1} \ K^{-1}$)

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

For the reaction,$2 NH_{3(g)} + CO_{2(g)} \rightleftharpoons NH_2CONH_{2(aq)} + H_2O_{(l)}$,find the value of the equilibrium constant at $295 \ K$. Given,the standard Gibbs energy change at the given temperature is $13.9 \ kJ \ mol^{-1}$.

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