Question

(a) What is the effect of temperature on equilibrium constant of an exothermic and endothermic reaction?

(b) The value of K_p for the reaction–

CO₂(g) + C(s) ⇌ 2CO(g)

is 3.0 at 1000 K. If initially PCO₂ = 0.48 bar and PCO = 0 bar and pure graphite is present, calculate the equilibrium partial pressures of CO and CO₂.

Answer 1

(a) The equilibrium constant for an exothermic reaction (negative ∆H) increases as the temperature increases.

The equilibrium constant for an endothermic reaction (positive ∆H) increases as the temperature increases.

(b) For the reaction, K_p = 3.0

Let ‘x’ be the decrease in pressure of CO₂,

Then

K_p = \(\frac{P^2_{CO}}{P_{CO_2}}\)

3 = \(\frac{(2X)^2}{(0.48-X)}\)

or 4x² = 3(0.48 – x)

or 4x² = 3(0.48 – x)

or 4x² = 1.44 – 3x

or 4x² + 3x – 1.44 = 0

a = 4, b = 3, c = – 1.44

(For quadratic equation, ax² + bx + c = 0)

x = \( {(-b \pm \sqrt{b^2-4ac}) \over 2a}\)

= \( {(-3 \pm \sqrt{(3)^2-4\times4\times(-1.44)}) \over 2\times4}\)

= \({(-3\pm5.66)} \over8\)

= \(\big({{-3+5.66}\over8}\big)\) (as value of x cannot be negative hence we neglect that value)

= \(2.66\over8\) = 0.33

x = \(2.66\over8\) = 0.33

\(\therefore\) Pco = 2x = 2 × 0.33 = 0.66 bar

Pco₂ = 0.48 – x = 0.48 – 0.33 = 0.15 bar