Chapter 6, Problem 6.103QP

Interpretation Introduction

Interpretation:

The $\text{ΔU}\text{and}\text{ΔH}$ values for reaction occurs when $1.248g$ of ethanol burned in calorimeter at $298Kand36.97kJ$ have to be calculated.

Concept Introduction:

At constant pressure, the exchange of heat between the system and surroundings is called as Enthalpy and its state function.

Enthalpy can be given by the formula,

$\text{H=U+PV}$

For any process, the enthalpy change is given as,

$\text{ΔH=ΔU+Δ(PV)}$

If the pressure is held constant,

  $\text{ΔH=ΔU+PΔV}$

  $\begin{array}{c}\text{ΔU = ΔH - }\left(\text{n}\text{RT}\right)\\ \text{where,}\\ \text{R and}\text{T}\text{values}\text{are}\text{constant}\text{.}\\ \text{ΔU = ΔH - RTΔn}\\ \text{ΔH = ΔU + RTΔn}\\ where\text{PV=nRT}\end{array}$

Where,

  $H$=enthalpy

  $\text{ΔH}$= change in enthalpy

  $U$=internal energy

  $\text{ΔU}$=Change in internal energy

  $P$=Pressure

  $V$=Volume

  T= Temperature

  $\text{Δn = No}\text{. of moles of product gas - No}\text{. of moles of reactant gas}$