What is the overall energy change if 1,500 g of N2 are reacted to make ammonia?

Question

Ammonia (NH3) is made by the direct combination of H2 and N2 gases according to this

reaction: N2(g) + 3H2(g) → 2NH3(g) + 22.0 kcal

What is the overall energy change if 1,500 g of N2 are reacted to make ammonia?

Answer 1

4.9*10³ kJ (explanation and both solution methods below)

This high reaction energy makes sense due to the high stability of standard state diatomic gases N₂ and H₂ (this applies for molecules such as N, O, F, Cl, Br, and I). Diatomic just means it makes a molecule by bonding 2 atoms of itself together. Standard states can apply to other phases like liquid water and solid carbon graphite, but that's beyond the scope of this problem.

Assuming we don't know how to convert kcal into kJ, we are going to solve this problem using enthalpies. (If you want to see the version with converting kcal to kJ, scroll down to the large text break).

We need to find the total change in enthalpy (aka change in H) by using the formula ΔH= H_f(products)-H_f(reactants), where Δ means "change in" and H_f means "enthalpy of formation." Enthalpy of formation is the enthalpy value you'll find associated to a molecule if you google its enthalpy.

Substituting in our molecules, we get:

ΔH=H_f(2NH3(g))-H_f(3H2(g))-H_f(N2(g)).

Let's go through each molecule:

H2(g): H2(g) has a H_f equal to zero because it is in its standard state as a diatomic gas.

N2(g): N2(g) has a H_f equal to zero because it is in its standard state as a diatomic gas.

NH3(g): 46.0 kJ/mol according to google.

Before we move on, I want to take a look back to the formula for a second. We have:

ΔH= H_f(products)-H_f(reactants).

If we substitute in the reactant values we found, we get ΔH= H_f(products) - 0kJ - 0kJ, or just

ΔH= H_f(products). This is important to note for any future chemistry problem you receive, as any reaction asking for reaction enthalpy with all standard state reactants will have a reaction enthalpy equal to the formation enthalpies of its products alone.

Now back to the problem at hand. We have the enthalpies figured out, but no idea much of each molecule is in the reaction. Since NH3(g) is the only molecule with a non-zero standard enthalpy of formation, we are concerned with finding the amount of moles of NH3(g).

Let's look back to the reaction and what we have been given already:

N2(g) + 3H2(g) → 2NH3(g), 1500g of N2(g).

Looking at the equation coefficients, there are 2 moles of NH3(g) produced for every 1 mole of N2(g) used. So, if we find the moles of N2(g), we find the moles of NH3(g). Luckily, since we are given the grams of N2(g) used, we can find the moles used by dividing by the molar mass of N2. N2 has a molar mass of 28.0134 g/mol.

So, 1500g N2 / 28.0134g N2 /mol N2 = 53.5458 moles of N2(g)

53.5458 mol N2(g) * 2 mol NH3 / 1 mol N2 = 107.0916 mol NH3(g)

To get the enthalpy of reaction, knowing that ΔH=H_f(NH3(g)), all we have to do now is multiply the moles of NH3(g) by its H_f of 46.0 kJ/mol.

107.0916 mol NH3 * 46.0 kJ/mol NH3 = 4926.2136 kJ = 4.9*10³ kJ (2 sig figs in given 1500g of N2).

If we want to use the kcal to kJ conversion of 1 kcal=4.184kJ, the 22 kcal* 4.184kJ = 92.048 kJ

This product of +92.048kJ is produced for every mole of N2 used (or every 2 moles or NH3, either way works).

Since we are given that 1500 grams of N2(g) used, we can find the moles of N2 used by dividing the given mass by the molar mass of N2. N2 has a molar mass of 28.0134 g/mol.

So, 1500g N2 / 28.0134g N2 /mol N2 = 53.5458 moles of N2(g)

We can multiply the 53.5458 moles of N2 by 92.048 kJ to get the energy produced by 1500 g of N2:

53.5458*92.048 kJ = 4928.7837 kJ = 4.9*10³ kJ (2 significant figures in given 1500g of N2).