High School Physics

# Numerical problem β Calculating Internal Energy Change (π«U) 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 (π«U) for a chemical Reaction to make ammonia.

## Numerical problem from the Thermodynamics chapter β with solution

Question:

The reaction of nitrogen with hydrogen to make ammonia has ΞH = -92.2 kJ. What is the value of ΞUin 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)β2 NH3 (g)
ΞH = -92.2 kJ

### Solution (STEP BY STEP)

IDENTIFY THE DATA GIVEN AND THE UNKNOWN

Known Unknown

Change in enthalpy (ΞH = -92.2 kJ) Change in internal energy (ΞU) = ?

Pressure (P = 40.0 atm)

Volume Change (ΞV = -1.12 L)

STRATEGY

We are given an enthalpy change ΞH, a volume change ΞV, and a pressure P and asked to find an energy change ΞE.

Rearrange the equation ΞH = ΞU+ PΞV to the form ΞU= ΞH β PΞV

and substitute the appropriate values for ΞH, P, and ΞV.

SOLVING

ΞU= ΞH β PΞV

where ΞH = -92.2 kJ

PΞV = (40.0 atm)(-1.12 L) = -44.8 L atm

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

ΞU= ΞH β PΞV = (-92.2 kJ) β (-4.52 kJ) = -87.7 kJ