Consider the following equilibrium, for which\(K_{p}=0.0752\) at \(480^{\circ} \mathrm{C} :\) $$2 \mathrm{Cl}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 4 \mathrm{HCl}(g)+\mathrm{O}_{2}(g)$$ \begin{equation} \begin{array}{l}{\text { (a) What is the value of } K_{p} \text { for the reaction }} \\ {4 \mathrm{HCl}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{Cl}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) ?} \\ {\text { (b) } \mathrm{What} \text { is the value of } K_{p} \text { for the reaction }} \\\ {\mathrm{Cl}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HCl}(g)+\frac{1}{2} \mathrm{O}_{2}(g) ?} \\ {\text { (c) What is the value of } K_{c} \text { for the reaction in part (b)? }}\end{array} \end{equation}

Step by step solution

01

Find Kp for the reverse reaction (4 HCl(g) + O2(g) ⇌ 2 Cl2(g) + 2 H2O(g))

To find Kp for the reverse reaction, we simply take the inverse of the given Kp value for the initial reaction. The inverse of Kp for the given reaction will be the Kp for the reverse reaction. $$K_p^{'} = \frac{1}{K_{p}} = \frac{1}{0.0752} = 13.298$$ So, the value of Kp for the reverse reaction is 13.298.

02

Find Kp for the simplified reaction (Cl2(g) + H2O(g) ⇌ 2 HCl(g) + 1/2 O2(g))

To find Kp for the simplified reaction, first notice that it is just half of the initial reaction. When we divide a reaction by a factor, the equilibrium constant is raised to the power of that factor. In this case, we are dividing the initial reaction by 2, so we need to take the square root of the initial Kp value. $$K_p^{''} = \sqrt{K_{p}} = \sqrt{0.0752} = 0.274$$ So, the value of Kp for the simplified reaction is 0.274.

03

Find Kc for the simplified reaction (Cl2(g) + H2O(g) ⇌ 2 HCl(g) + 1/2 O2(g))

To find the value of Kc for the simplified reaction, we first need to know the relationship between Kp and Kc. The two constants are connected by the following formula: $$K_{p} = K_{c}(RT)^{\Delta n}$$ Where R is the gas constant (0.08206 L*atm/mol*K), T is the temperature in Kelvin (480+273.15=753.15K), and Δn is the change in the number of moles of gas between reactants and products (in this case, Δn = 2 - 3 + 0.5 = -0.5). Using the Kp value for the simplified reaction, we can find Kc. $$K_{c} = \frac{K_{p}}{(RT)^{\Delta n}} = \frac{0.274}{(0.08206 \times 753.15)^{-0.5}} = 36.87 $$ So, the value of Kc for the simplified reaction is 36.87.

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