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

NTA Abhyas 2020 · Chemistry · Chemical Equilibrium

500 mL of 0.150 M AgNO 3 solution were added in 500 mL of 1.09 M Fe...

500 mL of 0.150 M AgNO3 solution were added in 500 mL of 1.09 M Fe2+ solution and the reaction is allowed to reach an equilibrium at 25∘C2 {5}^{∘} \text{C}.

Ag+(aq)+Fe2+(aq)⇌Fe3+(aq)+Ag(s){\text{Ag}}^{+} \left(\text{aq}\right) + {\text{Fe}}^{2 +} \left(\text{aq}\right) ⇌ {\text{Fe}}^{3 +} \left(\text{aq}\right) + \text{Ag} \left(\text{s}\right)

For 25 mL of the solution, 30 mL of 0.0832 M KMnO4 were required for oxidation. Calculate equilibrium constant (M-1) of the reaction at 25∘C2 {5}^{∘} \text{C}.
[Report your answer by rounding it upto nearset whole number]

Correct answer

3

Step-by-step solution

{Ag+(aq)+Fe2+(aq)⇌Fe3+(aq)+Ag(s)Initial75mM545mM00Reaction−46mM−46mM+46mM+46mM————————————————————Final29mM499mM46mM46mM\begin{cases} & {\text{Ag}}^{+} \left(\text{aq}\right) & + & {\text{Fe}}^{2 +} \left(\text{aq}\right) & ⇌ & {\text{Fe}}^{3 +} \left(\text{aq}\right) & + & \text{Ag} \left(\text{s}\right) \\ \text{Initial} & 7 5 \text{mM} & & 5 4 5 \text{mM} & & 0 & & 0 \\ \text{Reaction} & - 4 6 \text{mM} & & - 4 6 \text{mM} & & + 4 6 \text{mM} & & + 4 6 \text{mM} \\ & \text{---------------} & & \text{---------------} & & \text{---------------} & & \text{---------------} \\ \text{Final} & 2 9 \text{mM} & & 4 9 9 \text{mM} & & 4 6 \text{mM} & & 4 6 \text{mM}\end{cases}<br/><br/>m.eqsof<br /><br />m.eqs of{\text{KMnO}_{4} = 3 0 \times 0 \cdot 0 8 3 2 \times 5 = 1 2 \cdot 4 8 \text{m.eqs}<br/>(Zfactor=5)<br/><br/>m.eqsof<br />(Z factor =5)<br /><br />m.eqs of{\text{Fe}}^{2 +}<mrow><mrow>= 12\cdot48 \text{Meqs} = 12\cdot48 \text{mM}</mrow><br/>(Zfactor=1)<br/><br/>TotalmMof</mrow> <br />(Z factor= 1)<br /><br />Total mM of{\text{Fe}}^{2 +}leftleft= 12 \cdot 48 \times 4 0 = 4 9 9 \text{mM}<br/><br/>AtEquilibriumconcentrations<br/><br/>[Ag+]=29mM1000mL=0⋅029M<br /><br />At Equilibrium concentrations<br /><br />\left[{\text{Ag}}^{+}\right] = \dfrac{2 9 \text{mM}}{1 0 0 0 \text{mL}} = 0 \cdot 0 2 9 \text{M}

\left[{\text{Fe}}^{2 +}\right] = \dfrac{4 9 9 \text{mM}}{1 0 0 0 \text{mL}} = 0 \cdot 4 9 9 \text{M}<br/><br/>[Fe3+]=46mM1000mL=0⋅046M<br /><br />\left[{\text{Fe}}^{3 +}\right] = \dfrac{4 6 \text{mM}}{1 0 0 0 \text{mL}} = 0 \cdot 0 4 6 \text{M}

KC=[Fe3+][Ag+][Fe2+]=0⋅0460⋅499×0⋅029=3.18M−1{\text{K}}_{\text{C}} = \dfrac{\left[{\text{Fe}}^{3 +}\right]}{\left[{\text{Ag}}^{+}\right]\left[{\text{Fe}}^{2 +}\right]} = \dfrac{0 \cdot 046}{0 \cdot 499 \times 0 \cdot 029} = 3.18{\text{M}}^{- 1}
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