NTA Abhyas 2020
- Exam
- NTA Abhyas
- Year
- 2020
- Subject
- Chemistry
- Chapter
- Chemical Equilibrium
NTA Abhyas 2020 · Chemistry · Chemical Equilibrium
Calculate Δ _ r G for the reaction at 27^ o C H _ 2 (g) +2 Ag ^ +...
Calculate for the reaction at {\text{27}^{o} \text{C}
{Given : P}_{\text{H}2} = \text{0.5 bar}; \left[{Ag}^{+}\right] = {10}^{- 5} ;
\left[H^{+}\right] = 1 {0}^{- 3} M ; {Δ}_{f} G^{∘} \left[{Ag}^{+} \left(\text{aq}\right)\right] = \text{77.1 kJ/mol}
{Given : P}_{\text{H}2} = \text{0.5 bar}; \left[{Ag}^{+}\right] = {10}^{- 5} ;
\left[H^{+}\right] = 1 {0}^{- 3} M ; {Δ}_{f} G^{∘} \left[{Ag}^{+} \left(\text{aq}\right)\right] = \text{77.1 kJ/mol}
Options
- A
- B
- C
- DNone of these
Correct answer
C.
Step-by-step solution
{Δ}_{r} G^{∘} = 0 - \text{77.1} × 2 = - \text{154.2 kJ/mol}
\text{Q} = \dfrac{{\left[H^{+}\right]}^{2}}{{P}_{\text{H} 2} · {\left[{Ag}^{+}\right]}^{2}}
Δ \text{G} = {Δ}_{r} G^{∘} + \text{RT ln Q}
{Δ}_{r} \text{G} = - \text{154.2} + \dfrac{\text{8.314} × \text{300 ln} \left(2 × 1 {0}^{4}\right)}{1 0 0 0}
\text{Q} = \dfrac{{\left[H^{+}\right]}^{2}}{{P}_{\text{H} 2} · {\left[{Ag}^{+}\right]}^{2}}
Δ \text{G} = {Δ}_{r} G^{∘} + \text{RT ln Q}
{Δ}_{r} \text{G} = - \text{154.2} + \dfrac{\text{8.314} × \text{300 ln} \left(2 × 1 {0}^{4}\right)}{1 0 0 0}