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 Δ*H *= -92.2 kJ. What is the value of Δ*U*in 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 H_{2}(g)→2 NH_{3 }(g) **

**Δ**

*H*= -92.2 kJ### Solution (STEP BY STEP)

**IDENTIFY** THE DATA GIVEN AND THE UNKNOWN

**Known Unknown**

Change in enthalpy (Δ*H *= -92.2 kJ) Change in internal energy (Δ*U)* = ?

Pressure (*P *= 40.0 atm)

Volume Change (Δ*V *= -1.12 L)

**STRATEGY**

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

Rearrange the equation Δ*H *= Δ*U*+ *P*Δ*V *to the form Δ*U*= Δ*H *– *P*Δ*V*

and substitute the appropriate values for Δ*H*, *P*, and Δ*V*.

**SOLVING**

**Δ U= ΔH – PΔV**

where **Δ H = -92.2 kJ**

*P*Δ*V** *= (40.0 atm)(-1.12 L) = -44.8 L atm

= (-44.8 )x 1000×10^{-6} m^{3} (101 x 10^{3}) Pa = -4520 J =** -4.52 kJ**

**Δ U= ΔH – PΔV**

*= (*-92.2 kJ) – (-4.52 kJ) =

**-87.7 kJ**

**CHECK** OR VALIDATE THE ANSWER

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