Video: Effect of Changing Pressure and Concentration on the Rate of the Reaction of Carbon Monoxide (CO) Gas with Nitrogen Dioxide (NO₂) Gas

Carbon monoxide reacts with nitrogen dioxide to form carbon dioxide and nitric oxide, as shown. CO(g) + NO₂(g) ⟶ CO₂(g) + NO(g) The rate law for this reaction is rate = k[NO₂]² and the reactants have a constant temperature and total volume. (a) By what factor does the rate change if the pressure of NO₂ decreases from 0.500 atm to 0.250 atm? (b) By what factor does the rate change if the concentration of CO changes from 0.010 M to 0.030 M?

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Video Transcript

Carbon monoxide reacts with nitrogen dioxide to form carbon dioxide and nitric oxide, as shown. CO gas plus NO₂ gas react to form CO₂ gas plus NO gas. The rate law for this reaction is rate equals K times the concentration of NO₂ squared and the reactants have a constant temperature and volume. By what factor does the rate change if the pressure of NO₂ decreases from 0.500 atmospheres to 0.250 atmospheres?

For this reaction we are looking at a rate equation that is entirely dependent on the concentration of NO₂. In this part of the question, we’re being told that the partial pressure of NO₂ is decreasing from 0.5 to 0.25 atmospheres; that’s decreasing by a half. We can use the ideal gas law to figure out the relationship between pressure and concentration. We’ve already been told in the question that we’re dealing with a constant temperature and constant volume conditions. If we rearrange the ideal gas equation, we can see that pressure is equal to 𝑛 divided by 𝑉 times RT. 𝑛 divided by 𝑉 is another way to express the concentration. And the rest of the terms on that side are constants, R and T.

This means that pressure and concentration are directly proportional. This means that if we take our final pressure 𝑃 𝑓 and divide it by our initial pressure 𝑃 𝑖, we’ll get the same value as if we do the same with the concentrations. This means that our final concentration of NO₂ divided by our initial concentration is equal to 0.250 atmospheres divided by 0.500 atmospheres. This is equal to 0.5. So we’ve demonstrated if all else is equal, if we halve the pressure, we’re going to halve the concentration. We can apply this to this question by understanding that, by halving the pressure of NO₂, we’re also halving its concentration.

The question asks that we work out what factor the rate changes by if we halve the pressure, which means we have to take the final rate and divide it by the initial rate. We calculate both rates using our rate equation, K multiplied by the concentration of NO₂ squared. And then we can substitute in the value we know for the final concentration, 0.5 times the initial concentration.

So we have K multiplied by half of the initial concentration all squared divided by K times the initial concentration squared. This gives us the square of 0.5, which is 0.25 multiplied by K times the initial concentration squared all over K times the initial concentration squared. We can cancel common terms, giving us a final factor of 0.25 which can be expressed as quarter. So the factor by which the rate changes when we drop the pressure of NO₂ from 0.5 to 0.25 atmospheres is a quarter.

By what factor does the rate change if the concentration of CO changes from 0.010 molar to 0.030 molar?

We known from part a that the factor we need to determine for the rate change can be calculated by taking the final rate and dividing it by the initial rate. This gives us the rate constant k multiplied the final concentration of NO₂ squared divided by k times the initial concentration of NO₂ squared.

If you take a quick look, you may notice this question is a little bit of a trick because there is no carbon monoxide term in the rate equation. Increasing the concentration of carbon monoxide does not change the concentration of nitrogen dioxide. So the final concentration of NO₂ is the same as the initial concentration of NO₂. Substituting that into our equation gives equivalent terms on the top and the bottom. Our final factor is one. The concentration of carbon monoxide has no effect whatever on the reaction rate.

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