## hw problem 11.39

904475052_
Posts: 27
Joined: Fri Sep 25, 2015 3:00 am

### hw problem 11.39

Can someone please explain to me why you multiply both K values instead of adding them?

Destiny Dare 1D
Posts: 21
Joined: Fri Sep 25, 2015 3:00 am

### Re: hw problem 11.39

What you want to do first is write K as an expression for each given equation.

For $2 BrCl (g) + H_{2} (g) \rightleftharpoons Br_{2} (g) + 2 HCl (g)$ (Equation 1):

$K_{1} = \frac{[Br_{2}][HCl]^{2}}{[BrCl]^{2}[H_{2}]}$

For $H_{2} (g) + Cl_{2} (g) \rightleftharpoons 2HCl (g)$ (Equation 2):

$K_{2} = \frac{[HCl]^{2}}{[H_{2}][Cl_{2}]}$

For $2 BrCl (g) \rightleftharpoons Br_{2} (g) + Cl_{2} (g)$ (Equation 3):

$K_{3} = \frac{[Br_{2}][Cl_{2}]}{[BrCl]^{2}}$

Looking at the expressions, you will see that if you multiply $K_{2}$ by $K_{3}$, the $[Cl_{2}]$ from the denominator of $K_{2}$ will cancel out with the $[Cl_{2}]$ in the numerator of $K_{3}$. What you're left with after multiplying, is the exact equilibrium expression for equation 1. Therefore you can solve for $K_{1}$ by multiplying $K_{2}$ and $K_{3}$.

$K_{2} = 4.0 \times 10^{31}$
$K_{3} = 377$

$K_{1} = (4.0\times 10^{31})(377)= 1.5\times 10^{34}$

904475052_
Posts: 27
Joined: Fri Sep 25, 2015 3:00 am

### Re: hw problem 11.39

thanks!

Isabelle Bautista 3H
Posts: 23
Joined: Fri Sep 29, 2017 7:06 am

### Re: hw problem 11.39

How did you even begin this problem? How did you know that this was the way to solve it?

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