$$2\text{COF}_2(g) \rightleftharpoons \text{CO}_2(g) + \text{CF}_4(g)$$, with forward and backward rate constants $$k_f$$ and $$k_b$$. $$K = 2$$ for the reaction at $$1000^\circ\text{C}$$. If a 5L mixture contains 0.105 mole $$\text{COF}_2$$, 0.22 mol $$\text{CO}_2$$ and 0.055 mol $$\text{CF}_4$$ at $$1000^\circ\text{C}$$.<br>(a) Will the mixture be at equilibrium?<br>(b) If the gases are not at equilibrium, in which direction will a net reaction occur?<br>(c) What is the amount of each gas present when equilibrium is established.<br>(d) What are the rate equations for the forward and backward reactions?<br>(e) Compare the rate constants at (i) initial stage and (ii) equilibrium stage

PLACEHOLDER ANSWER - the question has five parts. (a) The volume cancels, so $$Q = \frac{0.22 \times 0.055}{0.105^2} \approx 1.10$$, which is not equal to $$K = 2$$, so the mixture is not at equilibrium. (b) Since $$Q < K$$, net reaction occurs in the forward direction. (c) If $$x$$ mol of $$\text{CO}_2$$ forms, $$\frac{(0.22 + x)(0.055 + x)}{(0.105 - 2x)^2} = 2$$, i.e. $$7x^2 - 1.115x + 0.00995 = 0$$, so $$x \approx 0.0095$$; at equilibrium there are about 0.086 mol $$\text{COF}_2$$, 0.2295 mol $$\text{CO}_2$$ and 0.0645 mol $$\text{CF}_4$$. (d) $$r_f = k_f[\text{COF}_2]^2$$ and $$r_b = k_b[\text{CO}_2][\text{CF}_4]$$. (e) The rate constants themselves do not change with time, and $$\frac{k_f}{k_b} = K = 2$$ at all stages; initially $$r_f > r_b$$, while at equilibrium $$r_f = r_b$$.