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Chemical Equilibrium question

2022 · 24 Jun · Shift 2 · Q16
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  5. /2022 · 24 Jun · Shift 2 · Q16

Chemical Equilibrium question

2022 · 24 Jun · Shift 2 · Q16

JEE MainChemistryChemical EquilibriumNumerical+4 / −1
PCl5PCl_5PCl5​ dissociates as PCl5PCl_5PCl5​(g) ⇌\rightleftharpoons⇌ PCl3PCl_3PCl3​(g) + Cl2Cl_2Cl2​(g) 5 moles of PCl5PCl_5PCl5​ are placed in a 200 litre vessel which contains 2 moles of N2N_2N2​ and is maintained at 600 K. The equilibrium pressure is 2.46 atm. The equilibrium constant Kp for the dissociation of PCl5PCl_5PCl5​ is ‾\underline{\hspace{2cm}}​×\times× 10 −-− 3. (nearest integer) (Given : R = 0.082 L atm K −-− 1 mol −-− 1; Assume ideal gas behaviour)
Numerical answer
View written solutionFree

Correct answer: 1107

  1. Let the degree of dissociation of PCl5PCl_5PCl5​ be xxx moles

Initially:

  • PCl5=5PCl_5 = 5PCl5​=5
  • PCl3=0PCl_3 = 0PCl3​=0
  • Cl2=0Cl_2 = 0Cl2​=0
  • N2=2N_2 = 2N2​=2 (inert gas)

At equilibrium:

  • PCl5=5−xPCl_5 = 5-xPCl5​=5−x
  • PCl3=xPCl_3 = xPCl3​=x
  • Cl2=xCl_2 = xCl2​=x
  • N2=2N_2 = 2N2​=2

So total moles at equilibrium are ntot=(5−x)+x+x+2=7+xn_{\text{tot}}=(5-x)+x+x+2=7+xntot​=(5−x)+x+x+2=7+x

  1. Use the ideal gas equation to find total equilibrium moles

Given:

  • P=2.46 atmP=2.46\ \text{atm}P=2.46 atm
  • V=200 LV=200\ \text{L}V=200 L
  • T=600 KT=600\ \text{K}T=600 K
  • R=0.082 L atm K−1mol−1R=0.082\ \text{L atm K}^{-1}\text{mol}^{-1}R=0.082 L atm K−1mol−1

ntot=PVRTn_{\text{tot}}=\frac{PV}{RT}ntot​=RTPV​

ntot=2.46×2000.082×600n_{\text{tot}}=\frac{2.46\times 200}{0.082\times 600}ntot​=0.082×6002.46×200​

ntot=49249.2=10n_{\text{tot}}=\frac{492}{49.2}=10ntot​=49.2492​=10

Thus, 7+x=10⇒x=37+x=10 \Rightarrow x=37+x=10⇒x=3

  1. Equilibrium moles

Therefore:

  • PCl5=5−3=2PCl_5 = 5-3=2PCl5​=5−3=2
  • PCl3=3PCl_3 = 3PCl3​=3
  • Cl2=3Cl_2 = 3Cl2​=3
  • N2=2N_2 = 2N2​=2

Total moles =10=10=10

  1. Find partial pressures

Partial pressure == = mole fraction ×\times× total pressure.

pPCl5=210×2.46=0.492 atmp_{PCl_5}=\frac{2}{10}\times 2.46=0.492\ \text{atm}pPCl5​​=102​×2.46=0.492 atm

pPCl3=310×2.46=0.738 atmp_{PCl_3}=\frac{3}{10}\times 2.46=0.738\ \text{atm}pPCl3​​=103​×2.46=0.738 atm

pCl2=310×2.46=0.738 atmp_{Cl_2}=\frac{3}{10}\times 2.46=0.738\ \text{atm}pCl2​​=103​×2.46=0.738 atm

  1. Write expression for KpK_pKp​

For PCl5(g)⇌PCl3(g)+Cl2(g)PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)PCl5​(g)⇌PCl3​(g)+Cl2​(g)

Kp=pPCl3 pCl2pPCl5K_p=\frac{p_{PCl_3}\,p_{Cl_2}}{p_{PCl_5}}Kp​=pPCl5​​pPCl3​​pCl2​​​

Substitute values:

Kp=(0.738)(0.738)0.492K_p=\frac{(0.738)(0.738)}{0.492}Kp​=0.492(0.738)(0.738)​

Kp=0.5446440.492=1.107≈1.107K_p=\frac{0.544644}{0.492}=1.107\approx 1.107Kp​=0.4920.544644​=1.107≈1.107

  1. Match with asked format

The question asks for KpK_pKp​ in the form ‾×10−3\underline{\hspace{1cm}}\times 10^{-3}​×10−3

So, 1.107=1107×10−31.107 = 1107\times 10^{-3}1.107=1107×10−3

Hence the required integer is 1107\boxed{1107}1107​

  1. Comparison with stored answer

Stored correct answer = 110711071107

This matches exactly.

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