Question

Write down the expressions for K p and Kc for the following reaction. How are
they related?
CaCO3 (s)<---->CaO (s) + CO2 (g)

Answer 1

Answer on Question #57590 – Chemistry – Other

Write down the expressions for $K_p$ and $K_c$ for the following reaction. How are they related?

$$\mathrm{CaCO_3} (s) < - - - > \mathrm{CaO} (s) + \mathrm{CO_2} (g)$$

Solution:

$$\mathrm{CaCO_3} (s) < - - - > \mathrm{CaO} (s) + \mathrm{CO_2} (g)$$

The equilibrium constant is given by

$$K_c' = \frac{[\mathrm{CaO}][\mathrm{CO_2}]}{[\mathrm{CaCO_3}]} \quad [\mathrm{CaCO_3}] = \text{constant} \quad [\mathrm{CaO}] = \text{constant}$$ $$K_c = [\mathrm{CO_2}] = K_c' \times \frac{[\mathrm{CaCO_3}]}{[\mathrm{CaO}]} \quad K_p = P_{\mathrm{CO_2}}$$

The concentration of solids and pure liquids are not included in the expression for the equilibrium constant.

The relationship between $K_c$ and $K_p$ is

$$K_p = K_c (\mathrm{RT})^{\Delta n} = K_c (0.0821T)^{\Delta n},$$

$\Delta n =$ moles of gaseous products – moles of gaseous reactants.

For $\mathrm{CaCO_3} (s) < - - - > \mathrm{CaO} (s) + \mathrm{CO_2} (g) \Delta n = 1 - 0 = 1$, we have

$$K_p = K_c (0.0821T).$$

Answer:

$$K_p = K_c (0.0821T).$$

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