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

NTA Abhyas 2020 · Chemistry · Chemical Equilibrium

Consider the following reactions in which all the reactants and the...

Consider the following reactions in which all the reactants and the products are in gaseous state.
2PQ⇌P2+Q2;K1=3.6×105\text{2PQ} ⇌ {P}_{2} + {Q}_{2}; {K}_{1} = 3.6 \times 1{0}^{5}
PQ+1/2R2⇌PQR;K2=6×10−3\text{PQ} + 1 / \text{2}{\text{R}}_{2} ⇌ \text{PQR}; {K}_{2} = 6 \times 1{0}^{- 3}
The value of K3{K}_{3} for the equilibrium
1/2P2+1/2Q2+1/2R2⇌PQR1 / \text{2}{\text{P}}_{2} + 1 / \text{2}{\text{Q}}_{2} + 1 / \text{2}{\text{R}}_{2} ⇌ \text{PQR} is

Options

  1. A3×1033 \times 1{0}^{3}
  2. B6×1036 \times 1{0}^{3}
  3. C1.0×10−51.0 \times 1{0}^{- 5}
  4. D5×10−35 \times 1{0}^{- 3}

Correct answer

C. 1.0×10−51.0 \times 1{0}^{- 5}

Step-by-step solution

Given , 2PQ⇌P2+Q2\text{2PQ} ⇌ {P}_{2} + {Q}_{2}
K1=[P2][Q2][PQ]2=3.6×105{K}_{1} = \dfrac{\left[{P}_{2}\right]\left[{Q}_{2}\right]}{[ PQ{]}^{2}} = 3.6 \times 1{0}^{5}
PQ+1/2R2⇌PQR;\text{PQ} + 1 / \text{2}{\text{R}}_{2} ⇌ \text{PQR};
K2=[PQR][PQ][R2]1/2{K}_{2} = \dfrac{[ \text{PQR} ]}{\left[\text{PQ}\right]{\left[{R}_{2}\right]}^{1 / 2}}
=6×10−3= 6 \times 1{0}^{- 3}
For the required equilibrium
1/2P2+1/2Q2+1/2R2⇌PQR;1 / \text{2}{\text{P}}_{2} + 1 / \text{2}{\text{Q}}_{2} + 1 / \text{2}{\text{R}}_{2} ⇌ \text{PQR};
K3=[PQR][P2]1/2[Q2]1/2[R2]1/2{K}_{3} = \dfrac{[ \text{PQR} ]}{{\left[{P}_{2}\right]}^{1 / 2}{\left[{Q}_{2}\right]}^{1 / 2}{\left[{R}_{2}\right]}^{1 / 2}}
K3=K2×1K1{K}_{3} = {K}_{2} \times \sqrt{\dfrac{1}{{K}_{1}}}
=6×10−3×13.6×105=6×10−30.6×103= 6 \times 1{0}^{- 3} \times \sqrt{\dfrac{1}{3.6 \times 1{0}^{5}}} = \dfrac{6 \times 1{0}^{- 3}}{0.6 \times 1{0}^{3}}
=1×10−5= 1 \times 1{0}^{- 5}
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