A vessel at 1000 K contains CO_2 with a pressure of 0.5 atm . Some of the CO_2 is converted into CO on the addition of graphite. If the total pressure - Chemistry MCQ FIITJEE AIEEE 2011 (Set Q) 2026 | ExamTest.live

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A vessel at 1000 K contains CO_{2} with a pressure of 0.5 atm . Some of the CO_{2} is converted into CO on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of K_{p} is:

Solution & Step-by-step Explanation

Reaction: CO_{2} (g) + C (s) ⇌ 2 CO (g) Initial pressure: P_{CO_{2}} = 0.5 atm, P_{CO} = 0 .At equilibrium: P_{CO_{2}} = 0.5 - x, P_{CO} = 2 x .Total pressure = (0.5 - x) + 2 x = 0.5 + x = 0.8 atm . x = 0.3 atm .Equilibrium partial pressures: P_{CO_{2}} = 0.5 - 0.3 = 0.2 atm . P_{CO} = 2 (0.3) = 0.6 atm . K_{p} = \frac{( P _{CO} ) ^{2}}{P _{CO_{2}}} = \frac{( 0.6 ) ^{2}}{0.2} = \frac{0.36}{0.2} = 1.8 atm .

Try it yourself before checking the explanation above.

A vessel at 1000 K contains CO_{2} with a pressure of 0.5 atm . Some of the CO_{2} is converted into CO on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of K_{p} is:

3 atm

0.3 atm

0.18 atm

1.8 atm

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