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Chemical Equilibrium question

2021 · 20 Jul · Shift 1 · Q16
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  5. /2021 · 20 Jul · Shift 1 · Q16

Chemical Equilibrium question

2021 · 20 Jul · Shift 1 · Q16

JEE MainChemistryChemical EquilibriumNumerical+4 / −1
2SO2SO_2SO2​(g) + O2O_2O2​(g) ⇌\rightleftharpoons⇌ 2SO3SO_3SO3​(g) In an equilibrium mixture, the partial pressures are PSO3PSO_3PSO3​ = 43 kPa; PO2PO_2PO2​ = 530 Pa and PSO2PSO_2PSO2​ = 45 kPa. The equilibrium constant KPKPKP = ‾\underline{\hspace{2cm}}​×\times× 10 −-− 2. (Nearest integer)
Numerical answer
View written solutionFree

Correct answer: 172

  1. Write the equilibrium expression

For 2SO2(g)+O2(g)⇌2SO3(g)2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)2SO2​(g)+O2​(g)⇌2SO3​(g)

the equilibrium constant in terms of partial pressures is Kp=(PSO3)2(PSO2)2(PO2)K_p=\frac{(P_{SO_3})^2}{(P_{SO_2})^2(P_{O_2})}Kp​=(PSO2​​)2(PO2​​)(PSO3​​)2​

  1. Convert all pressures to the same unit

Given:

  • PSO3=43 kPaP_{SO_3}=43\,\text{kPa}PSO3​​=43kPa
  • PSO2=45 kPaP_{SO_2}=45\,\text{kPa}PSO2​​=45kPa
  • PO2=530 Pa=0.530 kPaP_{O_2}=530\,\text{Pa}=0.530\,\text{kPa}PO2​​=530Pa=0.530kPa
  1. Substitute into the expression

Kp=(43)2(45)2(0.530)K_p=\frac{(43)^2}{(45)^2(0.530)}Kp​=(45)2(0.530)(43)2​

  1. Calculate numerator and denominator

432=184943^2=1849432=1849 452=202545^2=2025452=2025 2025×0.530=1073.252025\times 0.530=1073.252025×0.530=1073.25

So, Kp=18491073.25≈1.7229K_p=\frac{1849}{1073.25}\approx 1.7229Kp​=1073.251849​≈1.7229

  1. Match with the required form

The question asks for Kp=‾×10−2K_p=\underline{\hspace{2cm}}\times 10^{-2}Kp​=​×10−2

So let the blank be nnn. Then n×10−2=1.7229n\times 10^{-2}=1.7229n×10−2=1.7229 n=1.7229×102=172.29n=1.7229\times 10^2=172.29n=1.7229×102=172.29

Nearest integer: n=172n=172n=172

  1. Final answer

Kp=172×10−2K_p=172\times 10^{-2}Kp​=172×10−2

Hence the required integer is 172.

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