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
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
Change in enthalpy (ΔH = -92.2 kJ) Change in internal energy (ΔU) = ?
Pressure (P = 40.0 atm)
Volume Change (ΔV = -1.12 L)
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.
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
CHECK OR VALIDATE THE ANSWER
The sign of ΔU is similar in size and magnitude ΔH, which is to be expected because energy transfer as work is usually small compared to heat.