Reaction Conditions

Question

For the reaction in a sealed container at a constant temperature,

2NO(g)+Br2​(g)2NOBr(g)

the initial concentrations are 5.20 M NO, 7.20 M Br₂, and no NOBr. At equilibrium, [NOBr] = 1.35 M. 

1. Calculate the value of K under the reaction conditions.

Answer 1

A good way to start any problem is to ask the questions:

1. What are we looking for?
2. What does this mean?

We are looking for the value of K, which is the equilibrium constant, which is the ratio of the concentrations at equilibrium of the products compared to the reactants, each taken to the power of the coefficient in the given balanced equation.

That is, K = [NOBr]²/([NO]²[Br₂]), where [X] = the equilibrium concentration of “X”.

We already know that [NOBr] at equilibrium is equal to 1.35M, so we need to determine the equilibrium concentrations of the two reactants (NO and Br₂). We often find this in problems — we start by looking for one thing, and we need to find something else first.

Now we can ask the questions:

1. What are we given in the problem?
2. What is happening in the problem?

We are given the initial concentrations of the reactants, told that there is no product (NOBr) initially, and we know the concentration of the NOBr at equilibrium.

We also know that NO and Br₂ will react with each other to produce NOBr. If we knew the change in concentrations when NO and Br₂ reacted, we could subtract this from the initial amounts and get their equilibrium concentrations. These are what we need to solve the problem. 

A nice way to organize the numbers that are given is to use an ICE (Initial, Change, Equilibrium) table. A very important thing to know is that the “change” row (C) must be in the same ratio as the balanced equation (just like we see when first learning stoichiometry — the reaction proceeds according to the ratios of the coefficients in the balanced equation). 

We don’t know what this change is right now, but we do know the ratio is 2:1:2 because of the balanced equation. Because it is unknown, we consider this to be a variable, so I’ll use x like we often do in algebra.

Initially, we have 5.20M NO(g) and 7.20M Br₂(g) and no NOBr(g), and at equilibrium we have 1.35M

      2NO(g)  +    Br_2​(g) <===> 2NOBr(g)

I      5.20M       7.20M           0

C      -2x          -x            +2x

E    5.20M-2x      7.20M-x.      1.35M

     

Because the reactants react (as the name implies), their concentrations decrease, which is why we have “-2x” and “-x” for NO and Br₂, respectively. We make more product, giving us the “+2x” for NOBr.

Note that we could have also written 0 + 2x (or just 2x) in the “Equilibrium” row for NOBr. This means that 2x = 1.35M, or x = 0.675M.

So,

[NO] = 5.20M - 2(0.675M) = 3.85M

[Br₂] = 7.20M - (0.675M) = 6.525M (we will keep the extra sig fig for now)

These are what we are looking for!

Therefore, K = [1.35]²/([3.85]²[6.525]) = 0.01884 = 0.0188 to 3 sig figs.

Note: While the K expression is correct in terms of concentration, technically the ICE table should be displayed in terms of number of moles. However, given no reason not to, we can assume the sealed container is at constant volume. Because there is a direct relationship between number of moles and concentration when volume is constant, we can use concentrations in the ICE table.

This problem would be more difficult if the volume also changed (but the same general approach of the bulleted questions apply).