Question about today's lecture problem?

[Wilson Yeh 1L]

In today's lecture there was an example on nitrogen dioxide formation. For method 1, how did the first two reactions add up to form the third reaction? 1. N2 (g) + O2 (g) -> 2NO (g)
2. 2NO (g) + O2 (g) -> 2NO (g)
3. N2 (g) + 2O2(g) -> 2NO2 (g)

So my question is how did it get from 1 and 2 to 3?

[andrewr2H]

Well in the first reaction nitrogen and oxygen combine to form nitrogen oxide. Then in the second reaction the nitrogen oxide that was formed combines again with another oxygen to produce nitrogen dioxide, the final product.

[skalvakota2H]

Given that nitrogen dioxide formation is a multi-step reaction, this means that in order to achieve the overall reaction there will be intermediate steps. In this case, NO (g) is a product of the first reaction but a reactant of the second reaction; thus, NO acts as an intermediate, and since the amounts of NO are balanced in the two reactions, they can be cancelled.

With that, adding the two chemical equations means bringing down the reactants that could not be cancelled out as well as the products, and if there are multiple of the same molecules, the sum is the new coefficient. Thus, this yields us reaction 3, which is the overall equation for nitrogen dioxide formation.

[Jingyi Li 2C]

1. N2 (g) + O2 (g) -> 2NO (g)
2. 2NO (g) + O2 (g) -> 2NO2 (g) [it should be NO2 instead of NO]
3. N2 (g) + 2O2(g) -> 2NO2 (g)

So, when you add up the first and the second reactions, you get N2(g) + O2(g) + 2NO (g) + O2(g) -> 2NO(g) + 2NO2(g). Then you combine O2 and cancel 2NO on both sides to get reaction 3.