Chapter 9, Problem 9.111QA

Interpretation Introduction

To determine:

If heating or cooling is required when the two given reactions are carried out.

a) $H_2\left(g\right)+{Cl}_2\left(g\right)\to 2HCl\left(g\right)$

b) $2NaCl\left(s\right)+H_{2}S{O}_4\left(l\right)\to 2HCl\left(g\right)+{Na}_{2}S{O}_4\left(s\right)$

Answer to Problem 9.111QA

Solution:

a) $H_2\left(g\right)+{Cl}_2\left(g\right)\to 2HCl\left(g\right)$: cooling is required.

b) $2NaCl\left(s\right)+H_{2}S{O}_4\left(l\right)\to 2HCl\left(g\right)+{Na}_{2}S{O}_4\left(s\right):$ heating is required.

Explanation of Solution

1) Concept:

To determine whether heating or cooling is required for any reaction, we need to know whether the reaction is exothermic or endothermic. This can be found out from the enthalpy of reaction, i.e., heat of reaction $(∆H{°}_{rxn})$.

Exothermic reaction is a process in which energy is given out. Negative sign is assigned for a process if heat is given out. So, for an exothermic reaction, the heat of reaction is negative, i.e.,${∆H}_{rxn}^o<0$.

Endothermic reaction is a process in which heat is gained or required by the reaction. A positive value is assigned for a process, which gains heat. So, for an endothermic reaction, the heat of reaction is positive, i.e.,${∆H}_{rxn}^o>0$.

We can find the heat of reaction by using the standard heats of formation for reactants and products and the stoichiometry of the reaction.

${∆H}_{rxn}^o=\sum n_{products }{∆H}_{f products}^o-\sum n_{reactants }{∆H}_{f reactants}^o$

where

${∆H}_{f products}^o$= Standard heat of formation of products

${∆H}_{f reactants}^o$= Standard heat of formation of reactants

$n_{products }$= moles of products

$n_{reactants }$= moles of reactants

2) Formula:

${∆H}_{rxn}^o=\sum n_{products }{∆H}_{f products}^o-\sum n_{reactants }{∆H}_{f reactants}^o$

3) Given:

${∆H_f^o}_{HCl \left(g\right)} = -92.3 kJ/mol$

${∆H_f^o}_{H_2\left(g\right)} =0 kJ/mol$

${∆H_f^o}_{{Cl}_2\left(g\right)}=0 kJ/mol$

${∆H_f^o}_{H_{2}S{O}_4\left(l\right)} =-814 .0 kJ/mol$

${∆H_f^o}_{NaCl\left(s\right)}=-411.2 lK/mol$

${∆H_f^o}_{{Na}_{2}S{O}_4\left(s\right)}=-1387.1 kJ/mol$

(Standard values are taken from Appendix 4)

4) Calculation:

Calculate the heat of reaction for both the reactions using standard heat of formation.

a) $H_2\left(g\right)+{Cl}_2\left(g\right)\to 2HCl\left(g\right)$

${∆H}_{rxn}^o=\left[n_{HCl\left(g\right) }{∆H}_{f HCl\left(g\right)}^o\right]-\left[n_{H_2\left(g\right) }{∆H}_{f H_2\left(g\right)}^o+n_{{Cl}_2\left(g\right) }{∆H}_{f {Cl}_2\left(g\right)}^o\right]$

Plug the given values,

${∆H}_{rxn}^o=\left[2 mol \times \left(-92.3 kJ/mol\right)\right]-\left[1 mol \times 0 kJ/mol+1 mol \times 0 kJ/mol\right]$

${∆H}_{rxn}^o=-184.6 kJ$

The reaction is exothermic. So, heat is one of the products of this reaction. Therefore, in order to carry out this reaction (forward reaction), cooling is required.

b) $2NaCl\left(s\right)+H_{2}S{O}_4\left(l\right)\to 2HCl\left(g\right)+{Na}_{2}S{O}_4\left(s\right)$

${∆H}_{rxn}^o=\left[n_{HCl\left(g\right) }{∆H}_{f HCl\left(g\right)}^o+n_{{Na}_{2}S{O}_4\left(s\right) }{∆H}_{f {Na}_{2}S{O}_4\left(s\right)}^o\right]-\left[n_{NaCl\left(s\right) }{∆H}_{f NaCl\left(s\right)}^o+n_{H_{2}S{O}_4\left(l\right) }{∆H}_{f H_{2}S{O}_4\left(l\right)}^o\right]$

Plug the given values,

${∆H}_{rxn}^o=\left[2mol \times \left(-92.3 kJ/mol\right)+1 mol \times \left(-1387.1 kJ/mol\right)\right]-\left[2 mol \times \left(-411.2 lK/mol\right)+1 mol \times \left(-814 .0 kJ/mol\right)\right]$

${∆H}_{rxn}^o= -1571.7 kJ+1636.4 kJ$

${∆H}_{rxn}^o=64.7 kJ$

The reaction is endothermic. So, heat is one of the reactants of this reaction. Therefore, in order to carry out this reaction (forward reaction), heating is required.

Conclusion:

The sign of the standard heat of reaction indicates whether the given reaction is exothermic or endothermic. For an exothermic reaction, the forward reaction is carried out by cooling or decreasing the temperature. For an endothermic reaction, the forward reaction is carried out by heating or increasing the temperature.