A mixture containing initially 3.90 mole of NO(g) and 0.88 mole of CO2(g) was allowed to react in a flask of volume 1.00 L at a

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A mixture containing initially 3.90 mole of NO(g) and 0.88 mole of CO2(g) was allowed to react in a flask of volume 1.00 L at a

certain temperature according to the reaction CO2(g) + NO(g) ↔ CO(g) + NO2(g) At equilibrium 0.11 mole of CO2(g) was found present in the reaction mixture. Calculate the equilibrium constant Kc for the reaction at the temperature of the experiment.

Chemistry

1 answer:

Ymorist [56] · Answer 1 · 2022-02-03T08:33:01+03:00

Answer:

Equilibrium constant Kc for the reaction will be 1.722

Explanation:

O2(g)+NO(g)→CO(g)+ NO2(g)

0.88 3.9 --- ---

0.88x 3.9-x x x

GIVEN:

0.88X-X= 0.11

⇒ X=0.77

CO2(g)+NO(g) → CO(g) + NO2(g)

0.88 3.9 --- ---

0.88-x 3.9-x x x

= 3.13 0.77 0.77

=0.11

Kc = $\frac{[CO] *[NO2]} {[CO2]*[NO]}$

= ${{0.77}×0.77÷{{0.11×3.13}}$

= 1.722