# How do you determine the amount of heat (in kJ) given off when 1.26 xx 10^4 g of ammonia are produced according to the equation N_2(g) + 3H_2(g) -> 2NH_3(g), DeltaH°_(rxn) = -92.6 kJ/mol?

## Assume that the reaction takes place under standard-state conditions at 25°C.

Jun 26, 2017

Here's how you can do that.

#### Explanation:

The problem provides you with the thermochemical equation that describes the synthesis of ammonia from nitrogen gas and hydrogen gas.

${\text{N"_ (2(g)) + 3"H"_ (2(g)) -> 2"NH"_ (3(g))" "DeltaH_ "rxn"^@ = - "92.6 kJ mol}}^{- 1}$

Notice that the standard enthalpy change of reaction, $\Delta {H}_{\text{rxn}}^{\circ}$, is equal to $- {\text{92.6 kJ mol}}^{- 1}$. This tells you that when $1$ mole of ammonia is produced by the reaction, $\text{92.6 kJ}$ of heat are being given off.

Keep in mind that the minus sign attached to the standard enthalpy change of reaction signifies heat lost.

$\Delta {H}_{\text{rxn" = color(red)(-)"92.6 kJ"color(white)(.)color(blue)("mol"^(-1)) => "92.6 kJ"color(white)(.)color(red)("lost")color(white)(.)color(blue)("per mole")color(white)(.)"of NH}} _ 3$ $\text{produced}$

Your first step will be to convert the number of grams of ammonia to moles by using the compound's molar mass

1.26 * 10^4 color(red)(cancel(color(black)("g"))) * "1 mole NH"_3/(17.031color(red)(cancel(color(black)("g")))) = "739.83 moles NH"_3

So, if $\text{92.6 kJ}$ of heat are being given off when $1$ mole is produced, it follows that when $739.83$ moles are being produced, the reaction will give off

$739.83 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles NH"_3))) * "92.6 kJ"/(1color(red)(cancel(color(black)("mole NH"_3)))) = color(darkgreen)(ul(color(black)("68,500 kJ}}}}$

The answer is rounded to three sig figs.

Keep in mind that when $\text{68,500 kJ}$ of heat are being given off, the standard enthalpy change of reaction must carry a minus sign!

DeltaH_ ("rxn for 739.83 moles NH"_ 3)^@ = - "68,500 kJ"