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Chemical Equilibrium question

2023 · 10 Apr · Shift 2 · Q19
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Chemical Equilibrium question

2023 · 10 Apr · Shift 2 · Q19

JEE MainChemistryChemical EquilibriumNumerical+4 / −1
A(g)⇌2 B(g)+C(g)\mathrm{A}(g) \rightleftharpoons 2 \mathrm{~B}(g)+\mathrm{C}(g)A(g)⇌2 B(g)+C(g) For the given reaction, if the initial pressure is 450 mm Hg450 \mathrm{~mm} ~\mathrm{Hg}450 mm Hg and the pressure at time t\mathrm{t}t is 720 mm Hg720 \mathrm{~mm} ~\mathrm{Hg}720 mm Hg at a constant temperature T\mathrm{T}T and constant volume V\mathrm{V}V. The fraction of A(g)\mathrm{A}(\mathrm{g})A(g) decomposed under these conditions is x×10−1x \times 10^{-1}x×10−1. The value of xxx is ‾\underline{\hspace{2cm}}​ (nearest integer)
Numerical answer
View written solutionFree

Correct answer: 3

  1. Reaction and change in moles

Given: A(g)⇌2B(g)+C(g)\mathrm{A}(g) \rightleftharpoons 2\mathrm{B}(g)+\mathrm{C}(g)A(g)⇌2B(g)+C(g)

Let initially there be nnn moles of A\mathrm{A}A.

If the fraction decomposed is α\alphaα, then:

  • moles of A\mathrm{A}A left =n(1−α)= n(1-\alpha)=n(1−α)
  • moles of B\mathrm{B}B formed =2nα= 2n\alpha=2nα
  • moles of C\mathrm{C}C formed =nα= n\alpha=nα

So total moles at time ttt are: n(1−α)+2nα+nα=n(1+2α)n(1-\alpha)+2n\alpha+n\alpha = n(1+2\alpha)n(1−α)+2nα+nα=n(1+2α)

  1. Use pressure ratio

At constant temperature and constant volume, P∝nP \propto nP∝n

Hence, PtP0=n(1+2α)n=1+2α\frac{P_t}{P_0} = \frac{n(1+2\alpha)}{n} = 1+2\alphaP0​Pt​​=nn(1+2α)​=1+2α

Given: P0=450 mm Hg,Pt=720 mm HgP_0 = 450\ \text{mm Hg}, \qquad P_t = 720\ \text{mm Hg}P0​=450 mm Hg,Pt​=720 mm Hg

Therefore, 720450=1+2α\frac{720}{450} = 1+2\alpha450720​=1+2α 1.6=1+2α1.6 = 1+2\alpha1.6=1+2α 2α=0.62\alpha = 0.62α=0.6 α=0.3\alpha = 0.3α=0.3

  1. Match with required form

The fraction decomposed is given as: x×10−1x\times 10^{-1}x×10−1

Since α=0.3=3×10−1\alpha = 0.3 = 3\times 10^{-1}α=0.3=3×10−1 we get: x=3x=3x=3

  1. Final answer

The required nearest integer is: 3\boxed{3}3​

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