Question

The equilibrium constant of the reaction \(\mathrm{CO}+\frac{1}{2} \mathrm{O}_{2}\) \(\rightarrow \mathrm{CO}_{2}\) at \(1000 \mathrm{K}\) and 1 atm is \(K_{P_{1}} .\) Express the equilibrium constant of the following reactions at \(1000 \mathrm{K}\) in terms of \(K_{P_{1}}\): \((a) \quad \mathrm{CO}+\frac{1}{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{CO}_{2} \quad\) at \(3 \mathrm{atm}\) \((b)\) \(\mathrm{CO}_{2} \rightleftharpoons \mathrm{CO}+\frac{1}{2} \mathrm{O}_{2} \quad\) at \(1 \mathrm{atm}\) \((c) 2 \mathrm{H}_{2} \rightleftharpoons 4 \mathrm{H}\)at 1 atm \((d) \quad \mathrm{H}_{2}+2 \mathrm{N}_{2} \rightleftharpoons 2 \mathrm{H}+2 \mathrm{N}_{2}\) at 2 atm \((e)\) \(6 \mathrm{H} \rightleftharpoons 3 \mathrm{H}_{2} \) at \(4 \mathrm{atm}\)

Short answer

a) CO + 0.5 O2 → CO2 at 3 atm b) CO2 → CO + 0.5 O2 at 1 atm c) 2 H2 → 4 H at 1 atm d) H2 + 2 N2 → 2 H + 2 N2 at 2 atm e) 6 H → 3 H2 at 4 atm Answers: a) \(K_P'=K_{P_1}(\frac{3}{1})^{\frac{-1}{2}}\) b) \(K_{P_2} = \frac{1}{K_{P_1}}\) c) Cannot find in terms of \(K_{P_1}\) d) Cannot find in terms of \(K_{P_1}\) e) Cannot find in terms of \(K_{P_1}\)

Step-by-step solution

a) CO + 0.5 O2 → CO2 at 3 atm

To find the \(K_{P}\) for this reaction at 3 atm, we simply need to recognize that this reaction is the same as the original reaction, but the pressure changed from 1 atm to 3 atm. The equilibrium constant expression remains the same, but we must convert the constant from 1 atm to 3 atm. The relationship between pressure and constant is given by: \(K_P'=K_{P_1}(\frac{P'}{P})^{\Delta n}\), where \(P'=3\) atm, \(P=1\) atm, and \(\Delta n\) is the change in moles of the reaction (products - reactants). In this case, \(\Delta n = (1)-(1+\frac{1}{2})=\frac{-1}{2}\). Now, we can find the new \(K_P\) : \(K_P'=K_{P_1}(\frac{3}{1})^{\frac{-1}{2}}\). #a) Answer:# The equilibrium constant for the reaction CO + 0.5 O2 → CO2 at 3 atm is \(K_P'=K_{P_1}(\frac{3}{1})^{\frac{-1}{2}}\).

b) CO2 → CO + 0.5 O2 at 1 atm

This reaction is simply the reverse of the initial reaction. In this case, we only need to find the inverse of the initial equilibrium constant, which will yield the equilibrium constant for the reverse reaction (\(K_{P_2}\)). \(K_{P_2} = \frac{1}{K_{P_1}}\). #b) Answer:# The equilibrium constant for the reaction CO2 → CO + 0.5 O2 at 1 atm is \(K_{P_2} = \frac{1}{K_{P_1}}\).

c) 2 H2 → 4 H at 1 atm

This reaction is not related to the initial reaction, so we cannot find an equilibrium constant for it in terms of \(K_{P_1}\). #c) Answer:# We cannot find an equilibrium constant for the reaction 2 H2 → 4 H at 1 atm in terms of \(K_{P_1}\).

d) H2 + 2 N2 → 2 H + 2 N2 at 2 atm

This reaction is also not related to the initial reaction and involves different species. We cannot find an equilibrium constant for it in terms of \(K_{P_1}\). #d) Answer:# We cannot find an equilibrium constant for the reaction H2 + 2 N2 → 2 H + 2 N2 at 2 atm in terms of \(K_{P_1}\).

e) 6 H → 3 H2 at 4 atm

This reaction is not related to the initial reaction either, so we cannot find an equilibrium constant for it in terms of \(K_{P_1}\). #e) Answer:# We cannot find an equilibrium constant for the reaction 6 H → 3 H2 at 4 atm in terms of \(K_{P_1}\).