Chapter 9, Problem 9.113QA

Interpretation Introduction

To find:

a) Write a balanced chemical equation for the reaction that takes place in a cup.

b) Determine if $NaOH$ or $H_2{SO}_4$ is left in the cup when the reaction is over

c) Calculate the enthalpy change per mole of $H_{2}O$ produced in the reaction.

Answer to Problem 9.113QA

Solution:

a) Balanced equation for the reaction of $NaOH$ and $H_2{SO}_4$ is

$2 NaOH\left(aq\right)+H_2{SO}_4\left(aq\right)\to 2H_{2}O\left(l\right)+{Na}_2{SO}_4\left(aq\right)$

b) Both $NaOH$ and $H_{2}S{O}_4$ are present in stoichiometric amounts, so none of the reactants are left in the cup when the reaction is over.

c) Enthalpy change per mole of $H_{2}O$ produced in the reaction is $-57.1 kJ$

Explanation of Solution

1) Concept:

We know the concentrations, volumes, densities, and initial and final temperatures of two solutions – a strong acid and a strong base. We are asked to calculate the value of $∆H_{rxn}$ for the reaction in which they neutralize each other.

The density of both aqueous solutions is nearly the same as that of water, which confirms that they are dilute solutions with heat capacities that are essentially the same as that of water. Therefore, equation for $q_{calorimeter}$ can be used to find the heat associated with the neutralization reaction. The common densities also mean the total mass of the solution in grams is the same as their total volume in milliliters. To calculate $∆H_{rxn}$, we will need to calculate the number of moles of each reactant and divide the $q_{calorimeter}$ by the number of moles of limiting reactant.

2) Formula:

i) $M=\frac{n}{L}$

where $M$= molarity of solution, $n$= number of moles of solute, $V$= Volume of solution in Liters.

ii) $q_{calorimeter}=m{c}_p∆T$

where,

$q$= amount of energy

$m$= mass

$c_p$= specific heat (the amount of energy required to raise the temperature per unit mass by $1^{o}C$.)

$∆T$= change in temperature (Final temperature – Initial temperature)

iii) $q_{rxn}= -q_{calorimeter}$

iv) The relation between enthalpy change and energy transferred is

$q_{calorimeter}=n∆H$

where, $n$ is the number of moles.

v) $1L=100 mL$

3) Given:

i) Molarity of $NaOH=1.0 M$

ii) Volume of $NaOH=100.0 mL$

iii) Molarity of $H_2{SO}_4=1.0 M$

iv) Volume of $H_2{SO}_4=50.0 mL$

v) Initial temperature = ${22.3}^{o}C$

vi) Final temperature = ${31.4}^{o}C$

vii) Density of mixed solution is $1.00 g/mL$

viii) The specific heat of the mixed solution is $4.18 J/g^{o}C$.

4) Calculation:

a) Writing the balanced equation for the neutralization reaction between $NaOH$ and $H_2{SO}_4$:

$NaOH$ is a strong base, and $H_{2}S{O}_4$ is a strong acid. It is a neutralization reaction, so salt and water are the products for this reaction. The reaction can be written as

$NaOH\left(aq\right)+H_2{SO}_4\left(aq\right)\to H_{2}O\left(l\right)+{Na}_2{SO}_4\left(aq\right)$

Taking an inventory of atoms on both sides of reaction,

$1Na+3H+1S+5O\to 2Na+2H+1S+5O$

Add coefficient 2 in front of $NaOH$ and $H_{2}O,$

$2 NaOH\left(aq\right)+H_2{SO}_4\left(aq\right)\to 2H_{2}O\left(l\right)+{Na}_2{SO}_4\left(aq\right)$

This is a balanced equation.

b) Limiting reactant:

Moles of $NaOH$

Convert volume in $mL to L.$

$100.0 mL \times \frac{1 }{1000 mL}=0.1000 L$

Calculate the moles of $NaOH$.

$0.1000L\times \frac{1.0 mol}{L}=0.1000 mol NaOH$

Convert volume in $mL to L.$

$50.0 mL{ H}_2{SO}_4\times \frac{1 L}{1000 mL}=0.0500 mol H_2{SO}_4$

Mole ratio of $NaOH:{ H}_2{SO}_4$ is 2:1, so calculating moles of $H_{2}S{O}_4$ for $0.100 mol$$NaOH$ as

$0.1000 mol NaOH\times \frac{1{ H}_2{SO}_4}{2 NaOH}=0.0500 mol { H}_2{SO}_4$

i.e., for the reaction $0.1000 mol NaOH$, $0.0500 mol H_2{SO}_4$ are needed and we have $0.0500 mol H_{2}S{O}_4$. That means both $NaOH$ and ${ H}_2{SO}_4$ react completely; therefore, none of the reactants will be left in the cup when the reaction is over.

c) Calculating enthalpy change per mole of $H_{2}O$:

Total volume of solution = $100.0 mL + 50.0 mL = 150.0 mL$

Calculating the mass of solution using density:

$150.0 mL\times \frac{1 g}{1 mL}=150.0 g solution$

Calculating the energy required for the reaction:

$q_{lost}=-q_{gained}$

Since no heat is lost to surroundings,

$0=-q_{gained}$

$-q=m{c}_p∆T$

$-q=150.0 g\times 4.18 J/g^{o}C\times (31.4^{o}C-{22.3}^{o}C)$

$-q=5705.7 J$

$q=-5.71 k J$

$0.1000 mol H_{2}O$ is produced from the reaction of $0.1000 mol NaOH$ and $0.0500 mol { H}_2{SO}_4$ Therefore, the enthalpy change per mole of $H_{2}O$

$q=n∆H$

$∆H=\frac{q}{n}=\frac{-5.71 k J}{0.1 mol H_{2}O}=-57.1 kJ/mol H_{2}O$

Conclusion:

The limiting reactant is determined using stoichiometry of the reaction. Then from the energy transferred, enthalpy change per mole of water is calculated.