નો $K_{sp}$ ........ થશે.
$(R = 8.314\, J\, K^{-1}\,mol^{-1})$
$Ag_2CO_{3(s)} \rightleftharpoons 2Ag^+_{(aq)} + CO^{2-}_{3(s)}$
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 \((-11.09)\) \(\mathrm{Ksp}=8.0 \times 10^{-12}\)
$A\left( s \right) \rightleftharpoons B\left( g \right) + C\left( g \right);{K_{{p_1}}} = x\,at{m^2}$
$D\left( s \right) \rightleftharpoons C\left( g \right) + E\left( g \right);{K_{{p_2}}} = y\,at{m^2}$
જો બન્ને ઘન પદાર્થો એકી સાથે વિયોજિત થાય તો કુલ દબાણ કેટલું થશે?
$(2)$ $Cl_2$ નું નિર્માણ વધુ માત્રામાં થાય છે.
$(3)$ $SO_2$ ની સાંદ્રતા ઘટે છે અને $SO_2$ $Cl_2$ ની વધે છે