Question

The equilibrium constant for the \(\mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2} \rightleftharpoons\) \(\mathrm{H}_{2} \mathrm{O}\) reaction at 1 atm and \(1200 \mathrm{K}\) is \(K_{P \cdot}\) Use this information to determine the equilibrium constant for the following reactions: \((a)\) at 1 atm \(\quad \mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{H}_{2} \mathrm{O}\) \((b)\) at \(7 \mathrm{atm} \quad \mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{H}_{2} \mathrm{O}\) \((c)\) at \(1 \mathrm{atm} \quad 3 \mathrm{H}_{2} \mathrm{O} \rightleftharpoons 3 \mathrm{H}_{2}+\frac{3}{2} \mathrm{O}_{2}\) \((d)\) at 12 atm \(\quad 3 \mathrm{H}_{2} \mathrm{O} \rightleftharpoons 3 \mathrm{H}_{2}+\frac{3}{2} \mathrm{O}_{2}\)

Short answer

Question: Given the equilibrium constant (Kp) for the reaction \(\mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{H}_{2} \mathrm{O}\) at 1 atm and 1200 K, determine the equilibrium constants for the following reactions at the specified pressures: a) The original reaction at 1 atm b) The original reaction at 7 atm c) The reaction \(3 \mathrm{H}_{2} \mathrm{O} \rightleftharpoons 3 \mathrm{H}_{2}+\frac{3}{2} \mathrm{O}_{2}\) at 1 atm d) The reaction \(3 \mathrm{H}_{2} \mathrm{O} \rightleftharpoons 3 \mathrm{H}_{2}+\frac{3}{2} \mathrm{O}_{2}\) at 12 atm Answer: a) Kp for the original reaction at 1 atm remains the same as the given value. b) Kp for the original reaction at 7 atm also remains the same as the given value, as Kp is independent of pressure. c) For the reaction in part c, Kp can be determined using the relationship \(Kp_{c} = (Kp)^{3}\). d) Kp for the reaction in part d remains the same as for part c, as Kp is independent of pressure.

Step-by-step solution

Finding Kp for part (a)

Since part (a) asks for the equilibrium constant (Kp) of the same reaction at the same conditions, the Kp for part (a) will remain the same.

Finding Kp for part (b)

The reaction in part (b) is the same as the given reaction, but at a different pressure of 7 atm. As the equilibrium constant Kp is dimensionless, it is independent of pressure. Therefore, the Kp for the reaction in part (b) will also remain the same as the given value.

Finding Kp for part (c)

For part (c), we have a reaction that is a multiple of the given reaction: \(3 \mathrm{H}_{2} \mathrm{O} \rightleftharpoons 3 \mathrm{H}_{2}+\frac{3}{2} \mathrm{O}_{2}\). When the stoichiometric coefficients of a reaction are multiplied by a factor, let's call the factor 'n', the new Kp (Kp_new) can be determined using the relationship: $$Kp_ {new}=({Kp})^{n}$$ In this case, n = 3, and the given Kp should be raised to the power of 3 to find the new equilibrium constant for part (c): $$Kp_ {c}=({Kp})^{3}$$

Finding Kp for part (d)

For part (d), we are given the same reaction as part (c) but at a different pressure of 12 atm. As we have discussed in part (b), the equilibrium constant Kp is dimensionless and is independent of pressure. Therefore, the equilibrium constant for part (d) will remain the same as for part (c): $$Kp_{d} = Kp_{c}$$