Chapter 17, Problem 6STP

Interpretation Introduction

Interpretation: The enthalpy change at constant pressure is to be found.

Concept introduction:

The standard enthalpy of formation is the amount of energy released or consumed when one mole of a substance is formed under standard conditions from its pure elements.

Standard temperature and pressure conditions are 298 K and 1 atm.

The enthalpy of formation for an element in its elemental state or naturally occurring state will always be zero because it takes no energy to form a naturally occurring compound.

Answer to Problem 6STP

The enthalpy change for the conversion of $15.0\text{g}$ of ${\text{H}}_2\text{(g)}$ to $\text{HF(g)}$ is $\text{3988 kJ}$

Explanation of Solution

The given chemical equation is:

  ${\text{H}}_2{\text{(g)+F}}_{\text{2}}\text{(g)}\to \text{2HF(g) }\Delta \text{H= -536kJ}$

Thus, it is given that when 1 mol of ${\text{H}}_2$ , energy released is $\text{536kJ}$

Given the amount of ${\text{H}}_2$ = $15.0\text{g}$

Moles of ${\text{H}}_2$ are found by dividing the given amount of ${\text{H}}_2$ by the molecular mass of ${\text{H}}_2$ .

Moles of ${\text{H}}_2$ = $\frac{15.0\text{g}}{2.016{\text{gmol}}^{\text{-1}}}=7.44\text{mol}$

For the reaction of $\text{1 mol}$ of ${\text{H}}_2\text{(g)}$ , heat released is = $\text{536 kJ}$

Thus, for the burning of $\text{7}\text{.44 mol}$ , heat released is = $\frac{\text{536kJ}}{1\text{mol}}\times 7.44\text{mol=3988 kJ}$