Answer
(a) $$P_{O_2} = 4.0 \times 10^{-88}$$
(b) Since $Q_p \lt K_p$, the reaction will proceed to the right.
(c) $$K_c = 6.1 \times 10^{88} $$
Work Step by Step
(a) - Write the Kp expression:
$$K_p = \frac{1}{P_{O_2}}$$
- Substitute and find $P_{O_2}$
$$2.5 \times 10^{87} = \frac{1}{P_{O_2}}$$
$$P_{O_2} = \frac{1}{2.5 \times 10^{87}} = 4.0 \times 10^{-88}$$
(b)
- Calculate Qp
$$Q_p = \frac{1}{0.21} = 4.8$$
Since $Q_p \lt K_p$, the reaction will proceed to the right.
(c)
1. Calculate $\Delta n$ (n is the amount of moles of gases):
$$\Delta n = n_{products} - n_{reactants} = 0 - 1 = -1 $$
2 . Calculate Kc:
$$K_c = \frac{K_p}{(RT)^{\Delta n}} = \frac{( 2.5 \times 10^{87} )}{(0.0821 \times 298 )^{ -1 }}$$
$$K_c = 6.1 \times 10^{88} $$
