Chemistry: The Science in Context (Fifth Edition)
Chemistry: The Science in Context (Fifth Edition)
5th Edition
ISBN: 9780393614046
Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster, Stacey Lowery Bretz, Geoffrey Davies
Publisher: W. W. Norton & Company
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Chapter 5, Problem 5.112AP

(a)

Interpretation Introduction

Interpretation: The balanced chemical equation for the given reaction takes place in the Styrofoam cup is to be stated.

The amount of NaOH or H2SO4 left in the Styrofoam cup is to be predicted.

The enthalpy change per mole of H2SO4 in the given reaction is to be calculated.

Concept introduction: In a balanced chemical equation the number of atom in product and reactant side are equal.

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

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

The enthalpy change per mole of an aqueous solution in the reaction is calculated by the formula,

ΔH=m×s×ΔTn

To determine: The balanced chemical equation for the given reaction takes place in the Styrofoam cup.

(a)

Expert Solution
Check Mark

Answer to Problem 5.112AP

Solution

The balanced chemical equation for the given reaction is,

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

Explanation of Solution

Explanation

The reaction between NaOH(aq) and H2SO4(aq) is an example of neutralization reaction and written as,

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

In the above equation the number of sodium and hydrogen atoms are not balanced

Hence, it is not a balanced chemical equation. In balanced chemical equation the number of atom in product and reactant side are equal.

To balance a general equation the given steps are followed.

  • The first step is to balance the number of metals.
  • The second step is to balanced the number of oxygen and hydrogen.

In the above equation the sodium atoms present on right side are twice to the sodium atoms present on left side. The sodium atom is balanced by multiplying the NaOH base with two.

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

Now, the hydrogen atoms present on left side are twice to the hydrogen atoms present on right side. The hydrogen atom is balanced by multiplying the H2O molecule with two.

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

Now, it is balanced chemical equation.

(b)

Interpretation Introduction

To determine: The amount of NaOH or H2SO4 left in the Styrofoam cup.

(b)

Expert Solution
Check Mark

Answer to Problem 5.112AP

Solution

There is no amount of NaOH or H2SO4 left in the Styrofoam cup

Explanation of Solution

Explanation

Given

Volume of NaOH sample is 100.0mL .

Molarity (M) of NaOH is 1.0M .

Volume of H2SO4 sample is 50.0mL .

Molarity of H2SO4 is 1.0M .

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 (1)

Substitute the values of Molarity and volume of solution of NaOH in equation (1).

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 (1).

1mol/mL=n250mL×1000n2=501000moln2=0.05mol

Where,

  • n2 is number of moles of H2SO4 used for 50mL .

The number of moles of NaOH (n1) and H2SO4 (n2) are in the ratio of 2:1 . This means the two moles of NaOH react with one mole of H2SO4 .

The number of moles is exactly same in the case of its balanced chemical equation form as explained.

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

Hence, there is no amount of NaOH or H2SO4 left in the Styrofoam cup.

(c)

Interpretation Introduction

To determine: The enthalpy change per mole of H2SO4 in the given reaction

(c)

Expert Solution
Check Mark

Answer to Problem 5.112AP

Solution

The enthalpy change per mole of H2SO4 is +114.114kJ/mol_ .

Explanation of Solution

Explanation

Given

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

The volume of H2SO4 solution (VH2SO4) is 50.0mL .

The initial temperature (T1) of each solution before mixing is 22.3°C .

The final temperature (T2) of mixed solution is 31.4°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 in the reaction is calculated by the formula,

ΔH=mMixed×s×ΔTn2 (2)

Where,

  • ΔH is the change in enthalpy.
  • m is the mass of mixed solution.
  • s is the specific heat of mixed solution.
  • ΔT is the change in temperature.
  • n2 is the number of moles of H2SO4 .

The change in temperature (ΔT) is calculated by,

ΔT=T2T1 (3)

Where,

  • T2 is final temperature of mixed solution.
  • T1 is initial temperature of each solution before mixing.

Substitute the values of T2 and T1 in equation (3).

ΔT=31.4°C22.3°CΔT=9.1°C

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

VMixed=VNaOH+VH2SO4 (4)

Where,

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

Substitute the value of VNaOH and VH2SO4 in equation (4).

VMixed=100.0mL+50.0mL=150.0mL

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

mMixed=VMixed×ρMixed (5)

Where,

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

Substitute the value of VMixed and ρMixed in equation (5).

mMixed=150.0mL×1g/mL=150.0g

Substitute the value of mMixed , s , ΔT and n2 in equation (2).

ΔH=150g×4.18J/(g.°C)×9.1°C0.05mol=1,14,114J/mol

Conversion of J/mol to kJ/mol .

1J/mol=103kJ/mol1,14,114J/mol=1,14,114×103kJ/mol=114.114kJ/mol

In the statement it is given that, no heat is lost during this neutralization reaction. That means the final value of ΔH would be in positive.

Hence, the enthalpy change per mole of H2SO4 is +114.114kJ/mol_ .

Conclusion

  1. a. The balanced chemical equation for the given reaction is,

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

  2. b. There is no amount of NaOH or H2SO4 left in the Styrofoam cup.
  3. c. The enthalpy change per mole of H2SO4 is +114.114kJ/mol_

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Chapter 5 Solutions

Chemistry: The Science in Context (Fifth Edition)

Ch. 5.6 - Prob. 11PECh. 5.7 - Prob. 12PECh. 5.7 - Prob. 13PECh. 5.8 - Prob. 14PECh. 5.8 - Prob. 15PECh. 5.9 - Prob. 16PECh. 5 - Prob. 5.1VPCh. 5 - Prob. 5.2VPCh. 5 - Prob. 5.3VPCh. 5 - Prob. 5.4VPCh. 5 - Prob. 5.5VPCh. 5 - Prob. 5.6VPCh. 5 - Prob. 5.7VPCh. 5 - Prob. 5.8VPCh. 5 - Prob. 5.9QPCh. 5 - Prob. 5.10QPCh. 5 - Prob. 5.11QPCh. 5 - Prob. 5.12QPCh. 5 - Prob. 5.13QPCh. 5 - Prob. 5.14QPCh. 5 - Prob. 5.15QPCh. 5 - Prob. 5.16QPCh. 5 - Prob. 5.17QPCh. 5 - Prob. 5.18QPCh. 5 - Prob. 5.19QPCh. 5 - Prob. 5.20QPCh. 5 - Prob. 5.21QPCh. 5 - Prob. 5.22QPCh. 5 - Prob. 5.23QPCh. 5 - Prob. 5.24QPCh. 5 - Prob. 5.25QPCh. 5 - Prob. 5.26QPCh. 5 - Prob. 5.27QPCh. 5 - Prob. 5.28QPCh. 5 - Prob. 5.29QPCh. 5 - Prob. 5.30QPCh. 5 - Prob. 5.31QPCh. 5 - Prob. 5.32QPCh. 5 - Prob. 5.33QPCh. 5 - Prob. 5.34QPCh. 5 - Prob. 5.35QPCh. 5 - Prob. 5.36QPCh. 5 - Prob. 5.37QPCh. 5 - Prob. 5.38QPCh. 5 - Prob. 5.39QPCh. 5 - Prob. 5.40QPCh. 5 - Prob. 5.41QPCh. 5 - Prob. 5.42QPCh. 5 - Prob. 5.43QPCh. 5 - Prob. 5.44QPCh. 5 - Prob. 5.45QPCh. 5 - Prob. 5.46QPCh. 5 - Prob. 5.47QPCh. 5 - Prob. 5.48QPCh. 5 - Prob. 5.49QPCh. 5 - Prob. 5.50QPCh. 5 - Prob. 5.51QPCh. 5 - Prob. 5.52QPCh. 5 - Prob. 5.53QPCh. 5 - Prob. 5.54QPCh. 5 - Prob. 5.55QPCh. 5 - Prob. 5.56QPCh. 5 - Prob. 5.57QPCh. 5 - Prob. 5.58QPCh. 5 - Prob. 5.59QPCh. 5 - Prob. 5.60QPCh. 5 - Prob. 5.61QPCh. 5 - Prob. 5.62QPCh. 5 - Prob. 5.63QPCh. 5 - Prob. 5.64QPCh. 5 - Prob. 5.65QPCh. 5 - Prob. 5.66QPCh. 5 - Prob. 5.67QPCh. 5 - Prob. 5.68QPCh. 5 - Prob. 5.69QPCh. 5 - Prob. 5.70QPCh. 5 - Prob. 5.71QPCh. 5 - Prob. 5.72QPCh. 5 - Prob. 5.73QPCh. 5 - Prob. 5.74QPCh. 5 - Prob. 5.75QPCh. 5 - Prob. 5.76QPCh. 5 - Prob. 5.77QPCh. 5 - Prob. 5.78QPCh. 5 - Prob. 5.79QPCh. 5 - Prob. 5.80QPCh. 5 - Prob. 5.81QPCh. 5 - Prob. 5.82QPCh. 5 - Prob. 5.83QPCh. 5 - Prob. 5.84QPCh. 5 - Prob. 5.85QPCh. 5 - Prob. 5.86QPCh. 5 - Prob. 5.87QPCh. 5 - Prob. 5.88QPCh. 5 - Prob. 5.89QPCh. 5 - Prob. 5.90QPCh. 5 - Prob. 5.91QPCh. 5 - Prob. 5.92QPCh. 5 - Prob. 5.93QPCh. 5 - Prob. 5.94QPCh. 5 - Prob. 5.95QPCh. 5 - Prob. 5.96QPCh. 5 - Prob. 5.97QPCh. 5 - Prob. 5.98QPCh. 5 - Prob. 5.99QPCh. 5 - Prob. 5.100QPCh. 5 - Prob. 5.101QPCh. 5 - Prob. 5.102QPCh. 5 - Prob. 5.103APCh. 5 - Prob. 5.104APCh. 5 - Prob. 5.105APCh. 5 - Prob. 5.106APCh. 5 - Prob. 5.107APCh. 5 - Prob. 5.108APCh. 5 - Prob. 5.109APCh. 5 - Prob. 5.110APCh. 5 - Prob. 5.111APCh. 5 - Prob. 5.112APCh. 5 - Prob. 5.113APCh. 5 - Prob. 5.114APCh. 5 - Prob. 5.115APCh. 5 - Prob. 5.116APCh. 5 - Prob. 5.117APCh. 5 - Prob. 5.118APCh. 5 - Prob. 5.119APCh. 5 - Prob. 5.120APCh. 5 - Prob. 5.121APCh. 5 - Prob. 5.122APCh. 5 - Prob. 5.123APCh. 5 - Prob. 5.124APCh. 5 - Prob. 5.125APCh. 5 - Prob. 5.126APCh. 5 - Prob. 5.127APCh. 5 - Prob. 5.128APCh. 5 - Prob. 5.129APCh. 5 - Prob. 5.130APCh. 5 - Prob. 5.131APCh. 5 - Prob. 5.132APCh. 5 - Prob. 5.133APCh. 5 - Prob. 5.134APCh. 5 - Prob. 5.135APCh. 5 - Prob. 5.136APCh. 5 - Prob. 5.137APCh. 5 - Prob. 5.138APCh. 5 - Prob. 5.139APCh. 5 - Prob. 5.140APCh. 5 - Prob. 5.141APCh. 5 - Prob. 5.142APCh. 5 - Prob. 5.143APCh. 5 - Prob. 5.144AP
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