Chemistry
Chemistry
4th Edition
ISBN: 9780393919370
Author: Thomas R. Gilbert
Publisher: NORTON
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Chapter 5, Problem 5.113AP
Interpretation Introduction

Interpretation: The maximum measured temperature in the Styrofoam cup is to be predicted for the given data.

Concept introduction: The maximum temperature (T2) in the Styrofoam cup is calculated by the formula,

T2=ΔHNaOH×n1mMixed×s+T1

To determine: The maximum measured temperature in the Styrofoam cup for the given data.

Expert Solution & Answer
Check Mark

Answer to Problem 5.113AP

Solution

The maximum measured temperature in the Styrofoam cup is 33.27°C._

Explanation of Solution

Explanation

Given

The volume of NaOH solution (VNaOH) is 100.0mL .

The volume of H2SO4 solution (VH2SO4) is 65.0mL .

The initial temperature (T1) of both solution before mixing is 25.0°C .

The density (ρMixed) of mixed solution is 1.00g/mL .

The specific heat (s) of mixed solution is 4.18J/(g.°C) .

The enthalpy change per mole of H2SO4 (ΔHH2SO4) is +114.114kJ/mol .

The maximum temperature (T2) in the Styrofoam cup is calculated by the formula,

ΔHNaOH=mMixed×s×ΔTn1ΔT=ΔHNaOH×n1mMixed×sT2T1=ΔHNaOH×n1mMixed×sT2=ΔHNaOH×n1mMixed×s+T1 (1)

Where,

  • ΔHNaOH is change in enthalpy per mole of NaOH .
  • m is mass of mixed solution.
  • s is specific heat of mixed solution.
  • n1 is number of moles of NaOH .
  • T2 is maximum measured temperature.
  • T1 is initial temperature of both solution.

The number of moles of solute dissolved per liter of the solution is stated as Molarity.

M=Numberofmolesofsolute(n)Vol.ofsolution(V)inLitre

If the amount of solution is given in mL , then the formula of molarity is rewritten as,

M=Numberofmolesofsolute(n)Vol.ofsolution(V)inmL×1000 (2)

Substitute the values of Molarity and volume of solution of NaOH in equation (2).

1mol/mL=n1100mL×1000n1=1001000moln1=0.10mol

Where,

  • n1 is number of moles of NaOH used for 100mL .

Substitute the values of Molarity and volume of solution of H2SO4 in equation (2).

1mol/mL=n265mL×1000n2=651000moln2=0.065mol

Where,

  • n2 is number of moles of H2SO4 used for 65.0mL .

The balanced chemical equation for the given reaction is,

2NaOH(aq)+H2SO4(aq)Na2SO4(aq)+2H2O(l)

In the above balanced chemical equation, the number of moles of NaOH is twice to number of moles of H2SO4 . However, in actual the number of moles of NaOH (n1) are less than twice to the number of moles of H2SO4 . This means NaOH acts as a limiting reactant.

The enthalpy change per mole of H2SO4 (ΔHH2SO4) is +114.114kJ/mol . The enthalpy change per mole of NaOH (ΔHNaOH) is calculated by,

ΔHNaOH=ΔHH2SO42=114.114kJ/mol2ΔHNaOH=57.057kJ/mol

Conversion of kJ/mol to J/mol .

1kJ/mol=103J/mol57.057kJ/mol=57.057×103J/mol=57057J/mol

So, the value of ΔHNaOH in J/mol is 57057J/mol .

The volume of mixed solution (VMixed) is the sum of the volume of NaOH and H2SO4 .

VMixed=VNaOH+VH2SO4 (3)

Where,

  • VNaOH is volume of NaOH solution.
  • VH2SO4 is volume of H2SO4 solution.

Substitute the value of VNaOH and VH2SO4 in equation (3).

VMixed=100.0mL+65.0mL=165.0mL

The mass of the mixed solution (mMixed) is calculated by the formula,

mMixed=VMixed×ρMixed (4)

Where,

  • ρMixed is density of mixed solution.
  • VMixed is volume of mixed solution.

Substitute the value of VMixed and ρMixed in equation (4).

mMixed=165.0mL×1g/mL=165.0g

Substitute the value of mMixed , s , T1 ΔHNaOH and n1 in equation (1).

T2=ΔHNaOH×n1mMixed×s+T1T2=57057J/mol×0.10mol165.0g×4.18J/(g.°C)+25.0°C=5705.7J689.7J/°C+25.0°CT2=33.27°C_

Hence, the maximum measured temperature in the Styrofoam cup is 33.27°C._

Conclusion

The maximum measured temperature in the Styrofoam cup is 33.27°C._

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Chapter 5 Solutions

Chemistry

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