Calculate the standard Gibbs free energy chan | vedclass

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Question

(A) The relationship between standard Gibbs free energy change $\Delta G^o$ and the equilibrium constant $K_p$ is given by the equation: $\Delta G^o =

Vedclass · Accepted Answer

$162.74 \ kJ \ mol^{-1}$

Vedclass · Answer

(A) The relationship between standard Gibbs free energy change $\Delta G^o$ and the equilibrium constant $K_p$ is given by the equation: $\Delta G^o = -RT \ln K_p$.Given values are $R = 8.314 \ J \ K^{-1} \ mol^{-1}$,$T = 298 \ K$,and $K_p = 3 	imes 10^{-29}$.Substituting these values into the equation:$\Delta G^o = -(8.314 \ J \ K^{-1} \ mol^{-1}) 	imes (298 \ K) 	imes \ln(3 	imes 10^{-29})$.$\Delta G^o = -2477.572 	imes (\ln 3 + \ln 10^{-29})$.$\Delta G^o = -2477.572 	imes (1.0986 - 66.779)$.$\Delta G^o = -2477.572 	imes (-65.6804) \approx 162735 \ J \ mol^{-1}$.Converting to $kJ \ mol^{-1}$,we get $\Delta G^o \approx 162.74 \ kJ \ mol^{-1}$.

Calculate the standard Gibbs free energy change $\Delta G^o$ at $298 \ K$ for the conversion of oxygen to ozone,given by the reaction: $\frac{3}{2} O_{2(g)} \rightleftharpoons O_{3(g)}$. The equilibrium constant $K_p$ for this conversion is $3 \times 10^{-29}$.

Similar Questions

Consider the following reaction at $298 \ K$.
$\frac{3}{2} O_{2(g)} \rightleftharpoons O_{3(g)} ; K_{P} = 2.47 \times 10^{-29}$.
$\Delta_{r} G^{\ominus}$ for the reaction is $ . . . . . . \ kJ$. (Given $R = 8.314 \ J \ K^{-1} \ mol^{-1}$)

The correct relation is:

$2O_{3(g)} \rightleftharpoons 3O_{2(g)}$
At $300 \ K$,ozone is $50\%$ dissociated. The standard free energy change at this temperature and $1 \ atm$ pressure is $(-) \dots \ J \ mol^{-1}$ (Nearest integer).
[Given: $\ln 1.35 = 0.3$ and $R = 8.3 \ J \ K^{-1} \ mol^{-1}$ ]

For the reaction $CH_{4(g)} + H_{2(g)} \longrightarrow C_2H_{6(g)}$,$K_p = 3.356 \times 10^{17}$,calculate $\Delta G^{\circ}$ for the reaction at $298 \ K$.

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

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