# Question fe987

Mar 15, 2016

$\Delta {H}_{\text{rxn" = -"1486 kJ}}$

#### Explanation:

The idea here is that the problem provides you with the enthalpy change of reaction ,$\Delta {H}_{\text{rxn}}$, associated with the reaction of $\text{4.620 g}$ of ammonia, ${\text{NH}}_{3}$, and asks for the enthalpy change of reaction when $\textcolor{red}{4}$ moles of ammonia react.

The balanced chemical equation for this combustion reaction looks like this

$\textcolor{red}{4} {\text{NH"_text(3(g]) + 5"O"_text(2(g]) -> 4"NO"_text((g]) + 6"H"_2"O}}_{\textrm{\left(g\right]}}$

You know that reacting $\text{4.620 g}$ of ammonia will give off $\text{100.78 kJ}$ of heat. This is equivalent to saying that when $\text{4.620 g}$ of ammonia react, the enthalpy change of reaction is

$\Delta {H}_{\text{rxn" = -"100.78 kJ}}$

The minus sign used here symbolized heat lost.

Use ammonia's molar mass to determine how many moles of ammonia would give off this much heat when reacted with enough oxygen gas.

4.620 color(red)(cancel(color(black)("g"))) * "1 mole NH"_3/(17.0305color(red)(cancel(color(black)("g")))) = "0.27128 moles NH"_3

So, if this many moles of ammonia give off $\text{100.78 kJ}$ of heat, it follows that four moles will give off

4color(red)(cancel(color(black)("moles NH"_3))) * "100.78 kJ"/(0.27128color(red)(cancel(color(black)("moles NH"_3)))) = "1486 kJ"

Since this much heat is being given off, you can say that the enthalpy change of reaction for the combustion of $4$ moles of ammonia is equal to

DeltaH_"rxn 4 moles" = color(green)(|bar(ul(color(white)(a/a)-"1486 kJ"color(white)(a/a)|)))#

SIDE NOTE The actual value of the enthalpy change of reaction for the combusion of 4 moles of ammonia is

$\Delta {H}_{\text{rxn 4 moles" = -"1267.2 kJ}}$

so you might want to double-check the values given to you in the problem.