Determine \(K_{c}\) for the reaction \(\mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})+\) \(\mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NOCl}(\mathrm{g}),\) given the following data at \(298 \mathrm{K}\) $$\frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NO}_{2}(\mathrm{g}) \quad K_{\mathrm{p}}=1.0 \times 10^{-9}$$ $$\operatorname{NOCl}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NO}_{2} \mathrm{Cl}(\mathrm{g}) \quad K_{\mathrm{p}}=1.1 \times 10^{2}$$ $$\mathrm{NO}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NO}_{2} \mathrm{Cl}(\mathrm{g}) \quad K_{\mathrm{p}}=0.3$$

Step by step solution

01

Analyze the given reactions and required combined reaction

Observe the reactions, the goal is to combine them in a way that the resulting reaction matches the original reaction given. Mostly this process involves reversing some reactions and/or multiplying them by some factor such that the final combined reaction gets the same as the initial one.

02

Combine reactions

By reversing the first reaction and keeping the second one as it is, and adding half of the third reaction, the result is the original reaction:\n \(-1/2(1/2 \mathrm{N}_{2} + \mathrm{O}_{2} \rightarrow \mathrm{NO}_{2})\),\n \(+1(\mathrm{NOCl} + 1/2 \mathrm{O}_{2} \rightarrow \mathrm{NO}_{2} \mathrm{Cl})\),\n \(+1/2(\mathrm{NO}_{2} + 1/2 \mathrm{Cl}_{2} \rightarrow \mathrm{NO}_{2}\mathrm{Cl}),\),\n This forms our desired equation: \(\mathrm{N}_{2} + \mathrm{O}_{2} + \mathrm{Cl}_{2} \rightarrow 2\mathrm{NOCl}\)

03

Calculate new equilibrium constants

The equilibrium constant for the reverse of a reaction is the inverse of the original reaction, and when a reaction is multiplied by a factor n, its equilibrium constant is raised to the nth power. Thus, the new equilibrium constants are: \n \(-1/2(\mathrm{K}_{\mathrm{p1}'}= \frac{1}{(K_{p1}^{1/2})}) = \frac{1}{(1.0 \times 10^{-9})^{1/2}} = 10^9^{1/2} = 10^{4.5}\) \n \(+1(K_{p2}= 1.1 \times 10^{2}) = 1.1 \times 10^{2}\), \n \(+1/2(K_{p3'} = (0.3)^{1/2}) = (0.3)^{1/2} = 0.548\)

04

Multiply the equilibrium constants

When reactions are added together, their equilibrium constants multiply. Thus, the equilibrium constant for the original reaction is \(K_{c} = K_{\mathrm{p1}'} \times K_{p2} \times K_{p3'} = 10^{4.5} \times 1.1 \times 10^{2} \times 0.548 = 6 \times 10^{6}\)

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