# Question 44af5

Feb 5, 2016

Here's what I got.

#### Explanation:

First thing first - you're missing the standard enthalpy of atomisation, ${\Delta}_{a t} {H}^{\circ}$, of hydrogen gas, ${\text{H}}_{2}$, so I'll use the value listed here

http://cccbdb.nist.gov/ea1.asp

So, you know that you have

$\text{For N"_2: color(white)(a) Delta_text(at)H^@ = "1167 kJ/mol}$

$\text{For H"_2: color(white)(a) Delta_text(at)H^@ = "436 kJ/mol}$

$\text{For NH"_3: color(white)(a) Delta_text(at)H^@ = "1167 kJ/mol}$

Now, the standard enthalpy of atomisation tells you the enthalpy change that accompanies the breaking of all the bonds in a compound to produce isolated atoms under standard conditions.

In this regard, the standard enthalpy of atomisation is always positive, since it symbolizes energy required to break chemical bonds, which in turn is always an endothermic process.

The problem wants you to find the standard enthalpy of formation, $\Delta {H}_{f}^{\circ}$, for ammonia. The standard enthalpy of formation for a compound is always calculated for the formation of one mole of that compound from its constituent elements in their stable form.

The chemical equation that describes the formation of ammonia from hydrogen gas and nitrogen gas looks like this

$\textcolor{red}{\frac{1}{2}} {\text{N"_text(2(g]) + color(blue)(3/2)"H"_text(2(g]) -> "NH}}_{\textrm{3 \left(g\right]}}$

Now, think about what needs to happen in order for a mole of ammonia molecules to be formed. Looking at the equation, you can say that

• $\textcolor{red}{\frac{1}{2}}$ moles of nitrogen molecules must be broken down into isolated nitrogen atoms

• $\textcolor{b l u e}{\frac{3}{2}}$ moles of hydrogen gas must be broken down into isolated hydrogen atoms

• $1$ mole of nitrogen atoms and $3$ moles of hydrogen atoms must be brought together to form $1$ mole of ammonia molecules

Take a look at the standard enthalpy change of atomisation for ammonia. When all the chemical bonds in ammonia are broken to form isolated ions

${\text{NH"_text(3(g]) -> "N"_text((g]) + 3"H}}_{\textrm{\left(g\right]}}$

the enthalpy change of reaction is

$\Delta H = {\Delta}_{a t} {H}^{\circ}$

This means that when the reverse process takes place, i.e. when one mole of nitrogen atoms and three moles of hydrogen atoms form one mole of ammonia molecules, the enthalpy change of reaction will be

$\Delta {H}^{'} = - {\Delta}_{a t} {H}^{\circ} \to$ making bonds is an exothermic process

So, you will need to provide enough energy to break all the bonds for nitrogen gas and hydrogen gas

color(red)(1/2) color(red)(cancel(color(black)("moles N"_2))) * "942 kJ"/(1color(red)(cancel(color(black)("mole N"_2)))) = "471 kJ"

color(blue)(3/2) color(red)(cancel(color(black)("moles H"_2))) * "436 kJ"/(1color(red)(cancel(color(black)("mole H"_2)))) = "654 kJ"

However, when these atoms will bond together, heat will be given off, so for the calculation

1 color(red)(cancel(color(black)("mole NH"_3))) * "1167 kJ"/(1color(red)(cancel(color(black)("mole NH"_3)))) = "1167 kJ"

make sure that you do not forget to add a minus sign, which symbolizes heat given off, when taking the total enthalpy change for this reaction.

So, the overall enthalpy change for this process, i.e. breaking bonds on the reactants' side and making bonds on the products' side will be

DeltaH = overbrace(("471 kJ" + "654 kJ"))^(color(purple)("bond breaking")) + overbrace ((-"1167 kJ"))^(color(purple)("bond making")) = -"42 kJ"#

Since this represents standard enthalpy of formation, you will have

$\Delta {H}_{f}^{\circ} = \textcolor{g r e e n}{- \text{42 kJ/mol}}$

We didn't get the value $- \text{48 kJ/mol}$ because of the value used for the standard enthalpy change of atomisation for hydrogen gas.