Chapter 6, Problem 6.27QP

Interpretation Introduction

Interpretation:

$\text{ΔU}$ for the given reaction has to be calculated.

Concept Introduction:

The change in internal change of gaseous reaction is to assume ideal gas behaviour and constant temperature.  The formula can be given as,

$\begin{array}{l}\text{ΔU}\text{=}\text{ΔH-Δ}\left(\text{PV}\right)\\ \text{ΔU}\text{=}\text{ΔH-Δ}\left(\text{nRT}\right)\\ \text{ΔU}\text{=}\text{ΔH-RTΔn}\end{array}$

Here,

$\begin{array}{l}\text{ΔU}\text{=}\text{Change}\text{in}\text{Internal}\text{energy}\\ \text{ΔH}\text{=}\text{Change}\text{in}\text{heat}\\ \text{n}\text{=}\text{number}\text{of}\text{moles}\\ \text{R}\text{=}\text{Gas}\text{contant}\\ \text{T}\text{=}\text{Temperature}\end{array}$