JEE MAIN - Chemistry (2021 - 20th July Morning Shift - No. 16)
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)
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)
Answer
172
Explanation
On reaction, 2SO2(g) + O2(g) $$\rightarrow$$ 2SO3(g)
Given values are : pSO3 = 45kPa, pSO2 = 530 Pa = 0.53 kPa
pSO2 = 43 kPa
Now, $${K_p} = {{{{[{p_{S{O_3}(g)}}]}^2}} \over {{{[{p_{S{O_2}(g)}}]}^2} \times [{p_{{O_2}}}]}}$$
On putting given values, we get
$$ \Rightarrow {K_p} = {{{{(43)}^2}} \over {{{(45)}^2} \times 0.53}}$$
$$ = {{1849} \over {2025 \times 0.53}} = {{1849} \over {1073.25}}$$
$$ = 1.7228$$
$$ = {{1.7228 \times {{10}^2}} \over {{{10}^2}}} = 172.28 \times {10^{ - 2}} = 172$$
Hence, the equilibrium constant, Kp = 172.
Given values are : pSO3 = 45kPa, pSO2 = 530 Pa = 0.53 kPa
pSO2 = 43 kPa
Now, $${K_p} = {{{{[{p_{S{O_3}(g)}}]}^2}} \over {{{[{p_{S{O_2}(g)}}]}^2} \times [{p_{{O_2}}}]}}$$
On putting given values, we get
$$ \Rightarrow {K_p} = {{{{(43)}^2}} \over {{{(45)}^2} \times 0.53}}$$
$$ = {{1849} \over {2025 \times 0.53}} = {{1849} \over {1073.25}}$$
$$ = 1.7228$$
$$ = {{1.7228 \times {{10}^2}} \over {{{10}^2}}} = 172.28 \times {10^{ - 2}} = 172$$
Hence, the equilibrium constant, Kp = 172.
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