Verified Solution • 2025–26 Session moderate

QJEEM21Ch87b3: JEE Main 2021 Chemistry - Class 11 JEEM21Ch87b3 Solution: 2SO 2 (g) + O 2 (g) 2SO 3 (g) In an equilibrium mixture, the…

2SO2(g) + O2(g) ⇌\rightleftharpoons 2SO3(g)

In an equilibrium mixture, the partial pressures are

PSO3 = 43 kPa; PO2 = 530 Pa and PSO2 = 45 kPa. The equilibrium constant KP = ___________ ×\times 10 −- 2. (Nearest integer)

Step-by-Step Verified Solution

On reaction, 2SO2(g) + O2(g) →\rightarrow 2SO3(g)

Given values are : pSO3 = 45kPa, pSO2 = 530 Pa = 0.53 kPa

pSO2 = 43 kPa

Now, Kp=[pSO3(g)]2[pSO2(g)]2×[pO2]{K_p} = {{{{[{p_{S{O_3}(g)}}]}^2}} \over {{{[{p_{S{O_2}(g)}}]}^2} \times [{p_{{O_2}}}]}}

On putting given values, we get

⇒Kp=(43)2(45)2×0.53\Rightarrow {K_p} = {{{{(43)}^2}} \over {{{(45)}^2} \times 0.53}}

=18492025×0.53=18491073.25= {{1849} \over {2025 \times 0.53}} = {{1849} \over {1073.25}}

=1.7228= 1.7228

=1.7228×102102=172.28×10−2=172= {{1.7228 \times {{10}^2}} \over {{{10}^2}}} = 172.28 \times {10^{ - 2}} = 172

Hence, the equilibrium constant, Kp = 172.

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