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
and 
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