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Chemical Equilibrium question

2023 · 6 Apr · Shift 2 · Q20
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  5. /2023 · 6 Apr · Shift 2 · Q20

Chemical Equilibrium question

2023 · 6 Apr · Shift 2 · Q20

JEE MainChemistryChemical EquilibriumNumerical+4 / −1
The equilibrium composition for the reaction PCl3+Cl2⇌PCl5\mathrm{PCl}_{3}+\mathrm{Cl}_{2} \rightleftharpoons \mathrm{PCl}_{5}PCl3​+Cl2​⇌PCl5​ at 298 K298 \mathrm{~K}298 K is given below: [PCl3]eq=0.2 mol L−1,[Cl2]eq=0.1 mol L−1,[PCl5]eq=0.40 mol L−1\left[\mathrm{PCl}_{3}\right]_{\mathrm{eq}}=0.2 \mathrm{~mol} \mathrm{~L}^{-1},\left[\mathrm{Cl}_{2}\right]_{\mathrm{eq}}=0.1 \mathrm{~mol} \mathrm{~L}^{-1},\left[\mathrm{PCl}_{5}\right]_{\mathrm{eq}}=0.40 \mathrm{~mol} \mathrm{~L}^{-1}[PCl3​]eq​=0.2 mol L−1,[Cl2​]eq​=0.1 mol L−1,[PCl5​]eq​=0.40 mol L−1 If 0.2 mol0.2 \mathrm{~mol}0.2 mol of Cl2\mathrm{Cl}_{2}Cl2​ is added at the same temperature, the equilibrium concentrations of PCl5\mathrm{PCl}_{5}PCl5​ is ‾\underline{\hspace{2cm}}​×10−2 mol L−1\times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}×10−2 mol L−1 Given : Kc\mathrm{K}_{\mathrm{c}}Kc​ for the reaction at 298 K298 \mathrm{~K}298 K is 20
Numerical answer
View written solutionFree

Correct answer: 49

  1. Write the equilibrium expression

For the reaction

PCl3+Cl2⇌PCl5\mathrm{PCl_3 + Cl_2 \rightleftharpoons PCl_5}PCl3​+Cl2​⇌PCl5​

we have

Kc=[PCl5][PCl3][Cl2]K_c = \frac{[\mathrm{PCl_5}]}{[\mathrm{PCl_3}][\mathrm{Cl_2}]}Kc​=[PCl3​][Cl2​][PCl5​]​

Given equilibrium concentrations:

[PCl3]=0.2,[Cl2]=0.1,[PCl5]=0.40[\mathrm{PCl_3}] = 0.2,\quad [\mathrm{Cl_2}] = 0.1,\quad [\mathrm{PCl_5}] = 0.40[PCl3​]=0.2,[Cl2​]=0.1,[PCl5​]=0.40

So,

Kc=0.40(0.2)(0.1)=0.400.02=20K_c = \frac{0.40}{(0.2)(0.1)} = \frac{0.40}{0.02} = 20Kc​=(0.2)(0.1)0.40​=0.020.40​=20

This matches the given value.

  1. Add chlorine and find the new reaction quotient

Initially at equilibrium,

[Cl2]=0.1 mol L−1[\mathrm{Cl_2}] = 0.1\ \text{mol L}^{-1}[Cl2​]=0.1 mol L−1

Now 0.2 mol0.2\ \text{mol}0.2 mol of Cl2\mathrm{Cl_2}Cl2​ is added. Since concentration units are used directly in the problem, this means the concentration increases by 0.2 mol L−10.2\ \text{mol L}^{-1}0.2 mol L−1.

Therefore, just after addition:

[PCl3]=0.2,[Cl2]=0.1+0.2=0.3,[PCl5]=0.40[\mathrm{PCl_3}] = 0.2, \quad [\mathrm{Cl_2}] = 0.1 + 0.2 = 0.3, \quad [\mathrm{PCl_5}] = 0.40[PCl3​]=0.2,[Cl2​]=0.1+0.2=0.3,[PCl5​]=0.40

Since more reactant is added, equilibrium shifts to the right.

  1. Set up the ICE change

Let xxx mol L−1^{-1}−1 of PCl3\mathrm{PCl_3}PCl3​ and Cl2\mathrm{Cl_2}Cl2​ react to form xxx mol L−1^{-1}−1 of PCl5\mathrm{PCl_5}PCl5​.

Then at the new equilibrium:

[PCl3]=0.2−x[\mathrm{PCl_3}] = 0.2 - x[PCl3​]=0.2−x [Cl2]=0.3−x[\mathrm{Cl_2}] = 0.3 - x[Cl2​]=0.3−x [PCl5]=0.40+x[\mathrm{PCl_5}] = 0.40 + x[PCl5​]=0.40+x

Using Kc=20K_c = 20Kc​=20,

0.40+x(0.2−x)(0.3−x)=20\frac{0.40 + x}{(0.2 - x)(0.3 - x)} = 20(0.2−x)(0.3−x)0.40+x​=20
  1. Solve the equation
0.40+x=20(0.2−x)(0.3−x)0.40 + x = 20(0.2 - x)(0.3 - x)0.40+x=20(0.2−x)(0.3−x)

First expand:

(0.2−x)(0.3−x)=0.06−0.5x+x2(0.2 - x)(0.3 - x) = 0.06 - 0.5x + x^2(0.2−x)(0.3−x)=0.06−0.5x+x2

So,

0.40+x=20(0.06−0.5x+x2)0.40 + x = 20(0.06 - 0.5x + x^2)0.40+x=20(0.06−0.5x+x2) 0.40+x=1.2−10x+20x20.40 + x = 1.2 - 10x + 20x^20.40+x=1.2−10x+20x2

Bring all terms to one side:

20x2−11x+0.8=020x^2 - 11x + 0.8 = 020x2−11x+0.8=0

Multiply by 5:

100x2−55x+4=0100x^2 - 55x + 4 = 0100x2−55x+4=0

Using quadratic formula,

x=55±552−4(100)(4)200x = \frac{55 \pm \sqrt{55^2 - 4(100)(4)}}{200}x=20055±552−4(100)(4)​​ x=55±3025−1600200x = \frac{55 \pm \sqrt{3025 - 1600}}{200}x=20055±3025−1600​​ x=55±1425200x = \frac{55 \pm \sqrt{1425}}{200}x=20055±1425​​ 1425≈37.75\sqrt{1425} \approx 37.751425​≈37.75

Thus,

x=55−37.75200≈17.25200≈0.08625x = \frac{55 - 37.75}{200} \approx \frac{17.25}{200} \approx 0.08625x=20055−37.75​≈20017.25​≈0.08625

or

x=55+37.75200≈0.46375x = \frac{55 + 37.75}{200} \approx 0.46375x=20055+37.75​≈0.46375

The second value is impossible because 0.2−x0.2 - x0.2−x would become negative.

Hence,

x≈0.08625x \approx 0.08625x≈0.08625
  1. Find new equilibrium concentration of PCl5\mathrm{PCl_5}PCl5​
[PCl5]new=0.40+x=0.40+0.08625=0.48625 mol L−1[\mathrm{PCl_5}]_{\text{new}} = 0.40 + x = 0.40 + 0.08625 = 0.48625\ \text{mol L}^{-1}[PCl5​]new​=0.40+x=0.40+0.08625=0.48625 mol L−1

So,

[PCl5]new≈48.625×10−2 mol L−1[\mathrm{PCl_5}]_{\text{new}} \approx 48.625 \times 10^{-2}\ \text{mol L}^{-1}[PCl5​]new​≈48.625×10−2 mol L−1

Rounded suitably,

[PCl5]new=49×10−2 mol L−1[\mathrm{PCl_5}]_{\text{new}} = 49 \times 10^{-2}\ \text{mol L}^{-1}[PCl5​]new​=49×10−2 mol L−1
  1. Final answer

The required integer is:

49\boxed{49}49​

This agrees with the stored correct answer.

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