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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 atomization of Br_ 2 (g) according to the reaction :...

Consider the atomization of Br_{2}(g) according to the reaction : Br2(g)⇌2Br(g)Givenantiloge(−7.02)=9x10−4{Br}_{2} \left(\text{g}\right)⇌ 2 \text{Br}\left(\text{g}\right) {Given antiloge(-7.02)=9x10-4} If the heat of atomization of bromine gas is 82 KJ/mol ; standard molar entropies of Br(g) and Br_{2}(g) are 175 and 245.4 JK^{-1} respectively, calculate degree of dissociation when the total pressure is 40 atm at 500 K. (assume α<<1inyourcalculation)\alpha<<1 in your calculation)

Options

  1. A2.37×10−32.37 \times 1 {0}^{- 3}
  2. B3.20×10−43.20 × 1 {0}^{- 4}
  3. C4.30×10−54.30 × 1 {0}^{- 5}
  4. D3.60×10−23.60 × 1 {0}^{- 2}

Correct answer

A. 2.37×10−32.37 \times 1 {0}^{- 3}

Step-by-step solution

For Br2(g)⇌2Br(g){Br}_{2} \left(\text{g}\right)⇌ 2 \text{Br}\left(\text{g}\right)
ΔSreac∘=2×175−244⋅4=105.6JK−1\Delta {S}_{reac}^{∘} =2 \times1 7 5 - 2 4 4 \cdot4 = 105.6 \text{J} { K}^{- 1}
ΔHreac∘=82KJm−1\Delta {H}_{reac}^{∘} =8 2 \text{KJ} { m}^{- 1}
ΔGreac∘=ΔHreac∘−TΔS∘=82000−500×105.6\Delta {G}_{reac}^{∘} = \Delta {H}_{reac}^{∘} - \text{T} \Delta S^{∘} = 8 2 0 0 0 - 5 0 0 \times105.6
= 2 9 \cdot2 \times1 {0}^{3} \text{J} = - \text{RT} \text\ln {K}_{P}
\text\ln {K}_{P} = \dfrac{- 2 9 7 0 0}{8 \cdot 3 1 4 \times 5 0 0} = -7.02
∴KP=9×10−4∴ {K}_{P} = 9 \times {10}^{- 4}
KP=4α2P1−α2{K}_{P} = \dfrac{4{α}^{2} \text{P}}{1 - {α}^{2}}<br/>as<br />as\alpha\text{\alpha}<<1∴1−α2=1< < 1 ∴ 1 - {α}^{2} = 1
∴KP=4×α2∴ {K}_{P} = 4 \times {α}^{2}P\text{P}
∵P=40∵ \text{P}\text{=}\text{40}<br/><br />∴ 9 \times {10}^{- 4} =4 ⋅ {α}^{2} ⋅ 40<br/><br />∴ {α}^{2} = \dfrac{9}{160} \times {10}^{- 4}<br />=\dfrac{9}{16}$ \times {10}^{- 5}
$= 2.37 \times {10}^{- 3}
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