Question

You want to determine ΔH o for the reaction Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) To do so, you first determine the heat capacity of a calorimeter using the following reaction, whose ΔH o is known: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) ΔH o = −57.32 kJ(A) Calculate the heat capacity of the calorimeter from these data: Amounts used: 50.0 mL of 2.00 M HCl and 50.0 mL of 2.00 M NaOH Initial T of both solutions: 16.9°C Maximum T recorded during reaction: 30.4°C Density of resulting NaCl solution: 1.04 g/mL c of 1.00 M NaCl(aq): 3.93 J/g·KkJ/°C(B) Use the result from part (a) and the following data to determine ΔH o rxn for the reaction between zinc and HCl(aq): Amounts used: 100.0 mL of 1.00 M HCl and 1.3078 g of Zn Initial T of HCl solution and Zn: 16.8°C Maximum T recorded during reaction: 20.5°C Density of 1.0 M HCl solution: 1.015 g/mL c of resulting ZnCl2(aq): 3.95 J/g·KkJ/mol

Answer 1

Answer:

(A) The heat capacity of the calorimeter is therefore = −2.1428KJ÷13.5°C

= −0.1587KJ/°C

 

(B) ΔHo for the reaction Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) = –15.42KJ

Explanation:

Solution

 

Calculate the heat actually evolved.

                 q = mcΔt

 

Finding the mass of the reactants in grams we have.

 

Use density. (50 mL + 50 mL ) = 100 mL of solution.

 

100 mL X 1.04g/mL     = 104 grams of solution. (mass = Volume X Density)

                       

 

Find the temperature change.

 

       Δt =tfinal - tinitial = 30.4°C – 16.9°C = 13.5°C

 

    q = mcΔt

       = 104grams × 3.93J/g°C  × 13.5°C = 5.51772×103J

                                         

 

       = 5.51772 × 103 J

 

This is the heat lost in the reaction between HCl and NaOH, therefore q = -5.52 × 103 J.

 

this is an exothermic heat producing reaction.

 To calculate the total heat of the reaction or heat per mole we have

  

50.0 mL of HCl X 2.00 mol HCl /(1000 mL HCl ) = 0.100 mol HCl

                            

 

The same quantity of base, 0.100 mole NaOH, was used.

The energy per unit mole is given by

  

i.e. molar enthalpy = J/mol = -5.52 × 103J / 0.100 mol

            = -5.52 × 104 J/mol

            = -55177.2 J/mol

            = -55.177 kJ/mol

 

Therefore, the enthalpy change for the neutralization of HCl and NaOH, that is the enthalpy, heat, of reaction is ΔH = -55.177 kJ/mol

Heat absorbed by the calorimeter = −57.32kJ − 55.177 kJ = −2.1428KJ

The heat capacity of the calorimeter is therefore = −2.1428KJ÷13.5°C

= −0.1587KJ/°C

 

(B) For the ZnCl we have

 

Calculate the heat actually evolved.

                            q = mcΔt

 

Finding the mass of the reactants in grams we have.

 

Use density.  100 mL of solution of HCl

 

100 mL X 1.015g/mL        = 101.5 grams of solution. (mass = Volume X Density)

                       

 

Find the temperature change.

 

       Δt =tfinal - tinitial = 20.5°C – 16.8°C = 3.7 °C

 

    q = mcΔt

       = 101.5grams × 3.95J/g°C  × 3.7°C = 1483.422×103J

                                         

 

       = -1483.422×103J

 

This is the heat lost in the reaction between HCl and NaOH, therefore q = -1.483 × 103 J.

 

this is an exothermic heat producing reaction.

 To calculate the total heat of the reaction or heat per mole we have

  

100.0 mL of HCl X 1.00 mol HCl /(1000 mL HCl ) = 0.100 mol HCl

                            

 

 

The energy per unit mole is given by

  

i.e. molar enthalpy = J/mol = -1.483 × 103J / 0.100 mol

                                         = -1.483 × 104 J/mol

                                         = -14834.22 J/mol

                                         = -14.834 kJ/mol

 

Therefore, the enthalpy change for the neutralization of HCl and NaOH, that is the enthalpy, heat, of reaction is ΔH = -14.834 kJ/mol

ΔHo for the reaction Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)

= -14.834 kJ –(0.1587KJ/°C×3.7°C) = -15.42KJ

ΔHo for the reaction Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) = –15.42KJ

Answer 2

The heat capacity of the calorimeter is −0.1587KJ/°C.

ΔHo for the reaction $Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)$ = –15.42KJ

Calculations and Parameters:

To calculate the heat actually evolved would be

q = mcΔt

To find the mass of the reactants in grams we have

Use density. (50 mL + 50 mL )
= 100 mL of solution.

100 mL X 1.04g/mL = 104 grams of solution.

Recall, (mass = Volume X Density)

To find the temperature change.

Δt = $tfinal - tinitial$

= 30.4°C – 16.9°C

= 13.5°C

q = mcΔt

= 104grams × 3.93J/g°C  × 13.5°C = 5.51772×103J

= 5.51772 × 103 J

This is the heat lost in the reaction between HCl and NaOH,

Therefore q = -5.52 × 103 J.

This is an exothermic heat-producing reaction.

To calculate the total heat of the reaction or heat per mole we have

50.0 mL of HCl X 2.00 mol HCl /(1000 mL HCl )

= 0.100 mol HCl

The same quantity of base, 0.100 mole NaOH, was used.

The energy per unit mole is given by:

• i.e. molar enthalpy = J/mol
• = -5.52 × 103J / 0.100 mol
• = -5.52 × 104 J/mol
• = -55177.2 J/mol
• = -55.177 kJ/mol

Therefore, the enthalpy change for the neutralization of HCl and NaOH, that is the enthalpy, heat, of reaction, is ΔH

= -55.177 kJ/mol

Heat absorbed by the calorimeter

= −57.32kJ − 55.177 kJ

= −2.1428KJ

The heat capacity of the calorimeter is, therefore,

= −2.1428KJ÷13.5°C

= −0.1587KJ/°C

For the ZnCl we have

Calculate the heat actually evolved.

q = mcΔt

Find the mass of the reactants in grams we have.

Use density.  100 mL of a solution of HCl

100 mL X 1.015g/mL        

= 101.5 grams of solution.

(mass = Volume X Density)

To find the temperature change.

Δt = $tfinal - tinitial$

= 20.5°C – 16.8°C

= 3.7 °C

q = mcΔt

= 101.5grams × 3.95J/g°C  × 3.7°C

= 1483.422×103J

This is the heat lost in the reaction between HCl and NaOH, therefore

q = -1.483 × 103 J.

This is an exothermic heat-producing reaction.

Therefore, to calculate the total heat of the reaction or heat per mole we have

100.0 mL of HCl X 1.00 mol HCl /(1000 mL HCl )

= 0.100 mol HCl

Hence, the energy per unit mole is given by

i.e. molar enthalpy

= J/mol = -1.483 × 103J / 0.100 mol

= -1.483 × 104 J/mol

= -14834.22 J/mol

= -14.834 kJ/mol

Therefore, the enthalpy change for the neutralization of HCl and NaOH, that is the enthalpy, heat, of reaction, is ΔH = -14.834 kJ/mol

ΔHo for the reaction $Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)$

= -14.834 kJ –(0.1587KJ/°C×3.7°C)

= -15.42KJ

ΔHo for the reaction $Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)$

= –15.42KJ

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