Question

Suppose we start with CO2 and H2, both at a concentration of 0.100 M. When the system reaches equilibrium, what are the concentrations of products and reactants?

Answer 1

Answer on Question#50953 – Chemistry – Inorganic Chemistry

Suppose we start with CO2 and H2, both at a concentration of 0.100 M. When the system reaches equilibrium, what are the concentrations of products and reactants?

Solution:

Reaction:

$$\mathrm{CO_2} + \mathrm{H_2} = \mathrm{CO} + \mathrm{H_2O}$$

The equilibrium constant K:

$$aA + bB = cC + dD; \quad K = \frac{[C]^a [D]^d}{[A]^a [B]^b};$$

K – the equilibrium constant; [A] – the equilibrium concentration (mol/L);

In this case we don’t know the equilibrium constant $(K_c)$ and the equilibrium concentration of any reagent. Suppose that $K_c = 1$

Initial concentration H2 = initial concentration CO2 = 0.1 M

$$K = [\mathrm{CO}][\mathrm{H_2O}] / [\mathrm{CO_2}][\mathrm{H_2}];$$

$C = n/V$; C – molar concentration (mol/L); n – mole (mol); V – volume (L);

In this case $V = \text{const}$, so we can use «n» instead of «C» in this equilibrium:

$$K = [\mathrm{CO}][\mathrm{H_2O}] / [\mathrm{CO_2}][\mathrm{H_2}];$$

The amount of reacted H2 = x,

$[\mathrm{CO}]_{\mathrm{eq}} = [\mathrm{H_2O}]_{\mathrm{eq}} = x$ (according to the reaction n(reactants) : n(products) = 1 : 1);

$$[\mathrm{CO_2}]_{\mathrm{eq}} = [\mathrm{H_2}]_{\mathrm{eq}} = 0.1 - x;$$ $$K = [\mathrm{CO}][\mathrm{H_2O}] / [\mathrm{CO_2}][\mathrm{H_2}];$$ $$1 = x^2 / (0.1 - x)^2;$$ $$x = 0.05$$

The equilibrium concentration of H2O = $[\mathrm{CO}]_{\mathrm{eq}} = 0.05 \, \mathrm{M}$;

$$[\mathrm{CO_2}]_{\mathrm{eq}} = [\mathrm{H_2}]_{\mathrm{eq}} = 0.1 - x = 0.1 - 0.05 = 0.05 \, \mathrm{M}$$

**Answer:** If $K_c = 1$,

$$[\mathrm{CO_2}]_{\mathrm{eq}} = [\mathrm{H_2}]_{\mathrm{eq}} = [\mathrm{CO}]_{\mathrm{eq}} = [\mathrm{H_2O}]_{\mathrm{eq}} = 0.05 \, \mathrm{M}.$$

(If we know the signification of $K_c$ we can calculate the concentration of products and reactants according to this example).

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