Question

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Thae value of $\sf K_c$ is 4.24 at 800K for the reaction ,
$\underline{\bf CO_(g)+H_2O_(g)\leftrightharpoons CO_2_(g)+H_2_(g)}$

Calculate equilibrium concentrations of CO_2 ,H_2 and H_2O at 800K,if only H_2O and CO are present initially at concentrations of 0.1M each .

$\boxed{\bf Note:-}}$



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Answer 1

For the reaction,

$CO(g) + H_2 O(g) = CO_2(g) + H_2(g)$

Initial concentration:

0.1M , 0.1M , 0 , 0

Let 'x' mole per litre of each of theproduct be formed.

At equilibrium:

(0.1 – x)M , (0.1 – x)M , xM , xM

where x is the amount of Carbon dioxide and Hydrogen, at equilibrium.

Hence, equilibrium constant can be written as,

$K_c = \frac{x²}{(0.1 – x)²} = 4.24$

→ x² = 4.24 (0.01 + x² – 0.2x)

→ x² = 0.0424 + 4.24 x² - 0.848x

→ 3.24x² - 0.848x + 0.0424 = 0

a = 3.24, b = -0.848, c = 0.0424

(for quadratic equation ax² + bx+c=0)

$x = \frac{( - b \: ± \: \sqrt{ {b}^{2} - 4ac) } }{2a}$

$= > x = \frac{ - ( - 0.848 \: ± \: \sqrt{( - 0.848)^{2} - 4(3.24)(0.0424) } }{2 \times 3.24}$

$= > x = \frac{ - 0.848±0.4118}{6.48}$

$x_1 = \frac{0.848 - 0.4118}{6.48} = 0.067$

$x_2 = \frac{0.848 + 0.4118}{6.48} = 0.194$

Here, the value 0.194 should be neglected because it will give concentration of the reactant which is more than initial concentration.

∴ The equilibrium concentrations are :-

$[CO_2] [H_2] = x = 0.067M$

$[CO] [H_2 O] = 0.1 - 0.067 = 0.033M$