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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Chemistry
Chapter
Chemical Equilibrium

NTA Abhyas 2020 · Chemistry · Chemical Equilibrium

For the following reaction A g _ (aq) ^ + + C l _ (aq) ^ - → AgC l...

For the following reaction

Ag(aq)++Cl(aq)−→AgCl(s)A{g}_{\left(aq\right)}^{+} + C{l}_{\left(aq\right)}^{-} \to AgC{l}_{\left(s\right)}

Given: ΔGfo\Delta{G}_{f}^{o} , AgCl = - 112.44 kJ/mol; ΔGfo\Delta{G}_{f}^{o} Cl–{Cl}^{–} = - 130 kJ/mol ; ΔGfoAg+\Delta{G}_{f}^{o}A{g}^{+} = 75 kJ/mol
Report your answer by rounding it upto nearest whole number. The Ksp{K}_{sp} of AgCl is n×10−10n \times 1{0}^{- 10} .The value of 'n' is.

Correct answer

1

Step-by-step solution

Ag(aq)++Cl(aq)−→AgCl(s)A{g}_{\left(aq\right)}^{+} + C{l}_{\left(aq\right)}^{-} \to AgC{l}_{\left(s\right)}

ΔGfo\Delta{G}_{f}^{o} , AgCl = - 112.44 kJ/mol; ΔGfoCl–\Delta{G}_{f}^{o}{Cl}^{–} = - 130 kJ/mol ; ΔGfoAg+\Delta{G}_{f}^{o}A{g}^{+} = 75 kJ/mol

ΔGreactiono=ΔGfo\Delta{G}_{reaction}^{o} = \Delta{G}_{f}^{o} , AgCl- [ ΔGfoAg++ΔGfoCl–\Delta{G}_{f}^{o}A{g}^{+} + \Delta{G}_{f}^{o}{Cl}^{–} ]

= - 112.44 kJ-[75 kJ- 130 kJ]

= -57.44 kJ

-2.303RTlog Ksp=ΔGreactiono{K}_{sp} = \Delta{G}_{reaction}^{o}

logKsp=574402.303×8.314×298≅10log {K}_{sp} = \dfrac{57440}{2.303 \times 8.314 \times 298} ≅ 10

Ksp=1×10−10{K}_{sp} = 1 \times 1{0}^{- 10}

Hence n = 1
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