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*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)β2 NH_{3 }(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*+ *P*Ξ*V *to the form Ξ*U*= Ξ*H *β *P*Ξ*V*

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

**SOLVING**

**Ξ U= ΞH β PΞV**

where **Ξ H = -92.2 kJ**

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

= (-44.8 )x 1000Γ10^{-6} m^{3} (101 x 10^{3}) Pa = -4520 J =** -4.52 kJ**

**Ξ U= ΞH β PΞV**

*= (*-92.2 kJ) β (-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.