Question

The value of the equilibrium constant for the reaction 2COF2 (g) ↔ CO2 (g) + CF4 (g) is Keq = 2. If the reaction container currently holds 1 mole each of CO2 and CF4 and 0.5 mole of COF2, then what will happen?

Answer 1

Answer on Question #45022, Chemistry, Inorganic Chemistry

Question

The value of the equilibrium constant for the reaction $2\mathrm{COF}2(g) \leftrightarrow \mathrm{CO}2(g) + \mathrm{CF}4(g)$ is $\mathrm{Keq} = 2$. If the reaction container currently holds 1 mole each of CO2 and CF4 and 0.5 mole of COF2, then what will happen?

Solution

$$\begin{array}{l} 2 \mathrm{COF}_2(g) \leftrightarrow \mathrm{CO}_2(g) + \mathrm{CF}_4(g) \\ k(\text{forward reaction}) = c(\mathrm{COF}_2)^2 = 0.52 = 0.25 \\ k(\text{reverse reaction}) = c(\mathrm{CO}_2) \cdot c(\mathrm{CF}_4) = 1.1 = 1 \\ K_{\mathrm{eq}} = k(\text{reverse reaction}) / k(\text{direct reaction}) = [\mathrm{CO}_2] \cdot [\mathrm{CF}_4] / [\mathrm{COF}_2]^2 = 2 \\ \end{array}$$

But in the considered case:

$$k(\text{reverse reaction}) / k(\text{direct reaction}) = 1 / 0.25 = 4$$

Therefore the concentrations of $\mathrm{CO}_2$ and $\mathrm{CF}_4$ will decrease while the concentration of $\mathrm{COF}_2$ will increase.

Answer: the reaction will proceed in reverse direction.

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