# Numerical problem – Calculating Internal Energy Change (𝚫U) for a Reaction (Thermodynamics chapter)

In this post, we will solve a numerical problem from the Thermodynamics chapter. We will use the enthalpy and internal energy equation. The goal of this numerical is to calculate Internal Energy Change (𝚫U) for a chemical Reaction to make ammonia.

## Numerical problem from the Thermodynamics chapter – with solution

Question:

The reaction of nitrogen with hydrogen to make ammonia has ΔH = -92.2 kJ. What is the value of ΔUin kilojoules if the reaction is carried out at a constant pressure of 40.0 atm and the volume change is -1.12 L?

N2(g) + 3 H2(g)→2 NH3 (g)
ΔH = -92.2 kJ

### Solution (STEP BY STEP)

IDENTIFY THE DATA GIVEN AND THE UNKNOWN

Known Unknown

Change in enthalpy (ΔH = -92.2 kJ) Change in internal energy (ΔU) = ?

Pressure (P = 40.0 atm)

Volume Change (ΔV = -1.12 L)

STRATEGY

We are given an enthalpy change ΔH, a volume change ΔV, and a pressure P and asked to find an energy change ΔE.

Rearrange the equation ΔH = ΔU+ PΔV to the form ΔU= ΔH PΔV

and substitute the appropriate values for ΔH, P, and ΔV.

SOLVING

ΔU= ΔH PΔV

where ΔH = -92.2 kJ

PΔV = (40.0 atm)(-1.12 L) = -44.8 L atm

= (-44.8 )x 1000×10-6 m3 (101 x 103) Pa = -4520 J = -4.52 kJ

ΔU= ΔH – PΔV = (-92.2 kJ) – (-4.52 kJ) = -87.7 kJ