# What is the enthalpy change for the reaction of 1.000 x 10^2g of nitrogen with sufficient oxygen according to the equation: N2 + O2 -->2NO H=+180.6 kJ?

Oct 24, 2015

$\Delta {H}_{\text{rxn" = +"645 kJ}}$

#### Explanation:

Start by taking a look at the enthalpy change of reaction, $\Delta {H}_{\text{rxn}}$, fo this particular reaction

$\text{N"_text(2(g]) + "O"_text(2(g]) -> 2"NO"_text((g])" }$, $\Delta {H}_{\textrm{r x n}} = + \text{180.6 kJ}$

The value given to you for the enthalpy change of reaction corresponds to the reaction of one mole of nitrogen gas and one mole of oxygen gas, to form 2 moles of nitric oxide, $\text{NO}$.

That means that you can expect the enthalpy change of reaction to vary depending on the exact number of moles of the reactants and of moles of product produced by the reaction.

To determine how many moles of nitrogen gas you have in $1.00 \cdot {10}^{2} \text{g}$, use its molar mass

1.00 * 10^2color(red)(cancel(color(black)("g"))) * ("1 mole N"_2)/(28.0134color(red)(cancel(color(black)("g")))) = "3.5697 moles N"_2

Since no mention was made on how much oxygen gas you have, you can safely assume that it will be in excess, which implies that all the moles of nitrogen gas will take part in the reaction.

Well, if one mole of nitrogen gas corresponds to an enthalpy change of reaction of $+ \text{180.6 kJ}$, it follows that $\text{3.5697 moles}$ will correspond to an enthalpy change of reaction of

3.5697color(red)(cancel(color(black)("moles N"_2))) * "180.6 KJ"/(1color(red)(cancel(color(black)("mole N"_2)))) = "644.69 kJ"

Rounded to three sig figs, the answer will be

DeltaH_"rxn" = color(green)(+"645 kJ")