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Chemical Equilibrium question

2019 · 10 Jan · Shift 1 · Q20
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  5. /2019 · 10 Jan · Shift 1 · Q20

Chemical Equilibrium question

2019 · 10 Jan · Shift 1 · Q20

JEE MainChemistryChemical EquilibriumMCQ+4 / −1
The values of Kp/Kc for the following reactions at 300 K are, respectively : (At 300 K, RT = 24.62 dm3 atm mol–1) N2N_2N2​(g) + O2O_2O2​(g) ⇌\rightleftharpoons⇌ 2 NONONO(g) N2O4N_2O_4N2​O4​(g) ⇌\rightleftharpoons⇌ 2 NONONO(g) N2N_2N2​(g) + 3H2H_2H2​(g) ⇌\rightleftharpoons⇌ 2 NH3NH_3NH3​(g)
  1. A
    24.62 dm3 atm mol–1, 606.0 dm6 atm2 mol–2 1.65 ×\times× 10–3 dm–6 atm–2 mol2
  2. B
    1,4.1 ×\times× 10–2 dm–3 atm–1 mol, 606 dm6 atm2 mol–2
  3. C
    1,24.62 dm3 atm mol–1 , 1.65 ×\times× 10–3 dm–6 atm –2 mol2
  4. D
    1, 24.62 dm3 atm mol–1, 606.0 dm6 atm2 mol–2
View written solutionFree

Correct answer: C

  1. Use the relation between KpK_pKp​ and KcK_cKc​

For a gaseous reaction,

Kp=Kc(RT)ΔnK_p = K_c (RT)^{\Delta n}Kp​=Kc​(RT)Δn

Hence,

KpKc=(RT)Δn\frac{K_p}{K_c} = (RT)^{\Delta n}Kc​Kp​​=(RT)Δn

where

Δn=(moles of gaseous products)−(moles of gaseous reactants)\Delta n = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)}Δn=(moles of gaseous products)−(moles of gaseous reactants)

Given at 300 K300\,\text{K}300K,

RT=24.62 dm3 atm mol−1RT = 24.62\, \text{dm}^3\,\text{atm mol}^{-1}RT=24.62dm3atm mol−1
  1. Reaction 1:
N2(g)+O2(g)⇌2NO(g)N_2(g) + O_2(g) \rightleftharpoons 2NO(g)N2​(g)+O2​(g)⇌2NO(g)

Here,

  • gaseous products =2= 2=2
  • gaseous reactants =1+1=2= 1+1=2=1+1=2

So,

Δn=2−2=0\Delta n = 2-2=0Δn=2−2=0

Therefore,

KpKc=(RT)0=1\frac{K_p}{K_c} = (RT)^0 = 1Kc​Kp​​=(RT)0=1
  1. Reaction 2:
N2O4(g)⇌2NO2(g)N_2O_4(g) \rightleftharpoons 2NO_2(g)N2​O4​(g)⇌2NO2​(g)

(From the options, this is clearly the intended dissociation reaction.)

Here,

  • gaseous products =2= 2=2
  • gaseous reactants =1= 1=1

So,

Δn=2−1=1\Delta n = 2-1=1Δn=2−1=1

Thus,

KpKc=(RT)1=24.62 dm3 atm mol−1\frac{K_p}{K_c} = (RT)^1 = 24.62\, \text{dm}^3\,\text{atm mol}^{-1}Kc​Kp​​=(RT)1=24.62dm3atm mol−1
  1. Reaction 3:
N2(g)+3H2(g)⇌2NH3(g)N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)N2​(g)+3H2​(g)⇌2NH3​(g)

Here,

  • gaseous products =2= 2=2
  • gaseous reactants =1+3=4= 1+3=4=1+3=4

So,

Δn=2−4=−2\Delta n = 2-4=-2Δn=2−4=−2

Therefore,

KpKc=(RT)−2=1(24.62)2\frac{K_p}{K_c} = (RT)^{-2} = \frac{1}{(24.62)^2}Kc​Kp​​=(RT)−2=(24.62)21​

Now,

(24.62)2≈606.14(24.62)^2 \approx 606.14(24.62)2≈606.14

Hence,

KpKc≈1606.14≈1.65×10−3\frac{K_p}{K_c} \approx \frac{1}{606.14} \approx 1.65 \times 10^{-3}Kc​Kp​​≈606.141​≈1.65×10−3

with units

dm−6 atm−2 mol2\text{dm}^{-6}\,\text{atm}^{-2}\,\text{mol}^2dm−6atm−2mol2
  1. Collect the three values

Respectively, the values are:

1,24.62 dm3 atm mol−1,1.65×10−3 dm−6 atm−2 mol21,\quad 24.62\, \text{dm}^3\,\text{atm mol}^{-1},\quad 1.65 \times 10^{-3}\, \text{dm}^{-6}\,\text{atm}^{-2}\,\text{mol}^21,24.62dm3atm mol−1,1.65×10−3dm−6atm−2mol2

This matches Option C.

  1. Comparison with stored correct answer

Stored correct answer: C

Our derived answer: C

So, the derived answer agrees with the stored correct answer.

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