Question

At a certain temperature, Kc = 0.0500 and ∆H = +39.0 kJ for the reaction below, 2 MgCl2(s) + O2(g) → 2MgO(s) + Cl2(g) Calculate Kc for the reaction, MgO(s) + ½ Cl2(g) → MgCl2(s) + ½ O2(g) and indicate whether the value of Kc will be larger or smaller at a lower temperature.

Answer 1

Explanation:

Since, it is shown that the reaction has been reversed. Therefore, value of $K_{c}$ will become $\frac{1}{K_{c}}$.

Hence, new $K_{c'} = \frac{1}{K_{c}}$

                                      = $\frac{1}{0.0500}$

                                      = 20

Also, the number of moles of each reactant has been halved. So, $K_{c''}$ for the reaction $MgO(s) + \frac{1}{2}Cl2(g) → MgCl_{2}(s) + \frac{1}{2} O2(g)$ will also get halved.

Therefore,     $K_{c''}$  = $K_{c'}$ = $(20)^{0.5}$

                               = 4.47

As the value of $\Delta H$ is given as +39.0 kJ. So, it means that the reaction is endothermic in nature. So, energy of reactants will be more than the products. Hence, according to Le Chatelier's principle reaction will move in the forward direction.

As a result, $K_{c}$ will also increase with increase in temperature.