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 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 – PΞ”V

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

SOLVING

Ξ”U= Ξ”H – 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 – PΞ”V = (-92.2 kJ) – (-4.52 kJ) = -87.7 kJ

CHECK OR VALIDATE THE ANSWER

The sign of Ξ”U is similar in size and magnitude Ξ”H, which is to be expected because energy transfer as work is usually small compared to heat.

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