Wize University Chemistry Textbook > Equilibrium

Solving for K in Different Scenarios

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Manipulating Equilibrium Constants

Now we are going to take a look at how the value for K changes in different scenarios (when we reverse the reaction, multiply a reaction by a coefficient, and add two reactions together).

First, let's write the K expression for the reaction below:

PCl3(g)+Cl2(g)PCl5(g) PCl_3(g)+Cl_2(g)⇌PCl_5(g)

Kc=[PCl5][Cl2][PCl3]K_c=\frac{[PCl_5]}{\left[Cl_2\right]\left[PCl_3\right]}

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What happens to K when we reverse the reaction?

PCl5(g)PCl3(g)+Cl2(g)PCl_5(g)⇌PCl_3(g)+Cl_2(g)


1Kc=[Cl2][PCl3][PCl5]\frac{1}{K_c}=\frac{\left[Cl_2\right]\left[PCl_3\right]}{[PCl_5]}


Wize Tip
If you reverse a reaction, take the inverse of K to find Knew!


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What happens to K when we multiply the reaction by a factor?

2PCl3(g)+2Cl2(g)2PCl5(g)2PCl_3(g)+2Cl_2(g)⇌2PCl_5(g)


(Kc)2=([PCl5][PCl3][Cl2])2\left(K_c\right)^2=\left(\frac{\left[PCl_5\right]}{[PCl_3]\left[Cl_2\right]}\right)^{^2}



Wize Tip
When you multiply a reaction by a coefficient, take K to the power of that factor to solve for Knew!


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How do we solve for Koverall given K for reactions that are being added together?



2N2(g)+O2(g)2N2O(g)] Kc,1+2N2O(g)+3O2(g)4NO2(g) Kc,22N2(g)+4O2(g)4NO2(g) Koverall\def\arraystretch{1.5}\begin{array}{cc}2N_2(g)+O_2(g)⇌2N_2O(g)]\ K_{c,1} \\ +2N_2O(g)+3O_2(g)⇌4NO_2(g)\ K_{c,2}\\ \hline 2N_2(g)+ 4O_2(g)⇌4NO_2(g) \ K_{overall}\end{array}








Koverall = (Kc,1)(Kc,2)K_{overall}\ =\ (K_{c,1})*(K_{c,2})

Wize Tip
When adding two reactions together, multiply their K values together to solve for Koverall!

Practice: Solving For K

Consider the following equilibria:
2HCl(g)H2(g)+Cl2(g)Kc,12HCl_{(g)} \rightleftharpoons H_{2(g)} +Cl_2{(g)} \hspace{20pt} K_c,1
2H2S2(g)2H2(g)+S2(g)Kc,22H_2S_2{(g)} \rightleftharpoons 2H_{2(g)} +S_2{(g)} \hspace{20pt} K_c,2
The equilibrium constant for the reaction is:
2HCl(g)+12S2(g)H2S2(g)+Cl2(g)Kc,32HCl_{(g)} + \frac{1}{2} S_{2(g)} \rightleftharpoons H_2S_{2(g)}+Cl_{2(g)} \hspace{20pt} K_c,3
a)Kc3=Kc,1Kc,2K_c3 = \frac{K_c,1}{K_c,2}

b)Kc3=Kc,2Kc,1K_c3 = \frac{K_c,2}{K_c,1}

c)Kc3=2Kc,1Kc,2K_c3 = \frac{2K_c,1}{K_c,2}

d) Kc3=Kc,1Kc,212K_c3 = \frac{K_c,1}{K_c,2^\frac{1}{2}}

e)Kc3=Kc,22Kc,1K_c3 = \frac{K_c,2^2}{K_c,1}

Extra Practice