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Question

(B) The relationship between Gibbs free energy change,enthalpy change,and entropy change is given by $\Delta G^\circ = \Delta H^\circ - T\Delta S^\cir

Vedclass · Accepted Answer

-$50$

Vedclass · Answer

(B) The relationship between Gibbs free energy change,enthalpy change,and entropy change is given by $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$.Also,$\Delta G^\circ = -RT \ln K$.Given $T = 300 	ext{ K}$,$\Delta H^\circ = 28.40 	ext{ kJ mol}^{-1} = 28400 	ext{ J mol}^{-1}$,$K = 1.8 	imes 10^{-7}$,and $R = 8.3 	ext{ J K}^{-1} 	ext{ mol}^{-1}$.First,calculate $\ln K$: $\ln K = \ln(1.8 	imes 10^{-7}) = \ln(18 	imes 10^{-8}) = \ln(2 	imes 3^2) - 8 \ln 10 = \ln 2 + 2 \ln 3 - 8 \ln 10$.Using $\ln x = 2.3 \log x$,we have $\ln 2 = 2.3 	imes 0.30 = 0.69$,$\ln 3 = 2.3 	imes 0.48 = 1.104$,and $\ln 10 = 2.3$.So,$\ln K = 0.69 + 2(1.104) - 8(2.3) = 0.69 + 2.208 - 18.4 = -15.502$.Now,$\Delta G^\circ = -RT \ln K = -(8.3)(300)(-15.502) = 38599.98 	ext{ J mol}^{-1} \approx 38.60 	ext{ kJ mol}^{-1}$.Finally,$\Delta S^\circ = \frac{\Delta H^\circ - \Delta G^\circ}{T} = \frac{28400 - 38600}{300} = \frac{-10200}{300} = -34 	ext{ J K}^{-1} 	ext{ mol}^{-1}$.The nearest integer value is $-34$.

Similar Questions

For the following reaction at $50^\circ$ $C$ and at $2 \text{ atm}$ pressure, $2N_2O_5(g) \rightleftharpoons 2N_2O_4(g) + O_2(g)$. $N_2O_5$ is $50\%$ dissociated. The magnitude of standard free energy change at this temperature is $x$. $x = . . . . . . \text{ J mol}^{-1}$.

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

At $300 \ K$,the equilibrium constant for a reaction is $10$. The standard free energy change (in $kJ \ mol^{-1}$) for the reaction is

Find out $\ln K_{eq}$ for the formation of $NO_2$ from $NO$ and $O_2$ at $298 \ K$.
$NO_{(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons NO_{2(g)}$
Given:
$\Delta G^o_f (NO_2) = 52.0 \ kJ/mol$
$\Delta G^o_f (NO) = 87.0 \ kJ/mol$
$\Delta G^o_f (O_2) = 0 \ kJ/mol$

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