$Ag_2CO_{3(s)} \rightleftharpoons 2Ag^+_{(aq)} + CO^{2-}_{3(s)}$
in water at $25\, ^o C$ is $(R = 8.314\, J\, K^{-1}\,mol^{-1})$
We have Given, $\Delta G^{o}=+63.3 \,\mathrm{KJ}=63.3 \times 10^{3} \,\mathrm{J}$
Thus, substitute $\Delta G^{o}=63.3 \times 10^{3} \,\mathrm{J}$
$\mathrm{R}=8.314 \,\mathrm{JK}^{-1} \,\mathrm{mol}^{-1}$
and $T=298 \,\mathrm{K}[25+273 \,\mathrm{K}]$ into the above equation to get, $63.3 \times 10^{3}=-2.303 \times 8.314 \times 298 \log \mathrm{K}_{\mathrm{sp}}$
$\log \mathrm{Ksp}=-11.09$
$\mathrm{Ksp}=$ antilog $(-11.09)$
$\mathrm{Ksp}=8.0 \times 10^{-12}$
$\Delta \mathrm{G}^{o}$ is related to $\mathrm{K}_{\mathrm{sp}}$ by the equation, $\Delta G^{o}=-2.303 \mathrm{RT} \log \mathrm{K}_{\mathrm{sp}}$
We have Given, $\Delta G^{o}=+63.3\, \mathrm{KJ}=63.3 \times 10^{3}\, \mathrm{J}$
Thus, substitute $\Delta G^{o}=63.3 \times 10^{3} \,\mathrm{J}$
$\mathrm{R}=8.314 \,\mathrm{JK}^{-1}\, \mathrm{mol}^{-1}$
and $\mathrm{T}=298\, \mathrm{K}[25+273 \,\mathrm{K}]$ into the above equation to get, $63.3 \times 10^{3}=-2.303 \times 8.314 \times 298 \log \mathrm{K}_{\mathrm{sp}}$
$\log \mathrm{Ksp}=-11.09$
$\mathrm{Ksp}=$ antilog $(-11.09)$ $\mathrm{Ksp}=8.0 \times 10^{-12}$
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$n$ $l$ $m_1$ $m_2$
$(i)\,\,C\,({\rm{graphite}})\, + \,{O_2}{\kern 1pt} (g)\, \to \,C{O_2}\,(g);\,\Delta r{H^\circleddash} = x\,\,kJ\,mo{l^{ - 1}}$
$(ii)\,\,C\,({\rm{graphite}})\, + \,\frac{1}{2}{O_2}{\kern 1pt} (g)\, \to \,CO\,(g);\,\Delta r{H^\circleddash} = y\,\,kJ\,mo{l^{ - 1}}$
$(iii)\,\,CO\,(g)\, + \,\frac{1}{2}{O_2}{\kern 1pt} (g)\, \to \,C{O_2}\,(g);\,\Delta r{H^\circleddash} = z\,\,kJ\,mo{l^{ - 1}}$
Based on the above thermochemical equations, find out which one of the following algebraic relationships is correct?
