Chapter 4, Problem 4.9NP

Interpretation Introduction

Interpretation:

The standard enthalpy of formation of FeS₂(s) needs to be calculated at a temperature of 600°C

Concept Introduction:

The standard enthalpy change for a reaction ${\text{ΔH}}^{\text{0}}$ is given in terms of the difference in the standard enthalpy of formation of the products and that of reactants.

i.e. $\Delta {\text{H}}^0\text{ = }{\displaystyle \sum {\text{n}}_{\text{p}}{\text{ΔH}}_{\text{f}}^{\text{0}}}\text{(products) - }{\displaystyle \sum {\text{n}}_{\text{r}}{\text{ΔH}}_{\text{f}}^{\text{0}}}\text{(reactants) ------(1)}$

n_p and n_r are the number of moles of the products and reactants

The enthalpy of formation at temperature T is related to the standard enthalpy of formation at 298 K by the following equation:

  $\begin{array}{l}{\text{ΔH}}_{\text{f}}^{\text{0}}{\text{(T) = ΔH}}_{\text{f}}^{\text{0}}\text{(298K) + }{\displaystyle \underset{\text{298}}{\overset{\text{T}}{\int }}{\text{ΔC}}_{\text{p}}}\text{dT ------(2)}\\ {\text{where ΔC}}_{\text{p}}={\displaystyle \sum {\text{C}}_{\text{p}}}\text{(products)}-{\displaystyle \sum {\text{C}}_{\text{p}}}\text{(reactants) i}\text{.e}\text{. the difference in the heat}\\ \text{capacities of the products and reactants}\end{array}$