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JEE Mains · Chemistry · STD 11 - 5. Thermodynamics and thermochemistry

નીચેની કોષની પ્રક્રિયા ધ્યાનમાં લો: \(\mathrm{Cd}_{(s)}+\mathrm{Hg}_{2} \mathrm{SO}_{4(s)}+\frac{9}{5} \mathrm{H}_{2} \mathrm{O}_{(l)} \rightleftharpoons \mathrm{CdSO}_{4} \cdot \frac{9}{5} \mathrm{H}_{2} \mathrm{O}_{(s)}+2 \mathrm{Hg}_{(l)}\) \(25^{\circ} {C}\) પર \({E}_{\text {cell }}^{0}\)નું મૂલ્ય \(4.315\, {~V}\) છે. જો \(\Delta {H}^{\circ}=-825.2\, {~kJ} \,{~mol}^{-1}\), પ્રમાણિત એન્ટ્રોપી ફેરફાર \(\Delta {S}^{\circ}\) \({J} \,{K}^{-1}\)માં \(........\) છે. (નજીકના પૂર્ણાંકમાં) [આપેલ: ફેરાડે અચળાંક \( = 96487 \, {C} \, {mol}^{-1} \)]

  1. A \(0.25\)
  2. B \(2.5\)
  3. C \(250\)
  4. D \(25\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(25\)

Step-by-step Solution

Detailed explanation

\(\Delta {G}^{\circ}=-{nFE}^{\circ}=\Delta {H}^{\circ}-{T} \Delta {S}^{\circ}\) \(=\frac{\Delta {H}^{\circ}+{nFE}^{\circ}}{{T}}\) \(=\frac{\left(-825.2 \times 10^{3}\right)+(2 \times 96487 \times 4.315)}{298}\) \(=\frac{-825.2 \times 10^{3}+832.682 \times 10^{3}}{298}\)…
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