Answer
$$K_p = 0.105$$
$$K_c = 2.05 \times 10^{-3} $$
Work Step by Step
1. Write the expression for $K_p$:
$$K_p = P_{CO_2}$$
2. Since $CO_2$ is the only gas in the mixture: $P_{CO_2} = P= 0.105 \space atm$
$$K_p = 0.105$$
3. Calculate $\Delta n$ (n is the amount of moles of gases):
$$\Delta n = n_{products} - n_{reactants} = 1 - 0 = 1 $$
4. Convert the temperature to Kelvin:
$$T/K = 350 + 273.15 = 623.15 $$
5. Calculate Kc:
$$K_c = \frac{K_p}{(RT)^{\Delta n}} = \frac{( 0.105 )}{(0.0821 \times 623.15 )^{ 1 }}$$
$$K_c = 2.05 \times 10^{-3} $$
