Answer:
(A) The calculated heat capacity of the calorimeter equals = −2.1428KJ÷13.5°C
= −0.1587KJ/°C
(B) The standard enthalpy change (ΔHo) for the reaction Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) is –15.42KJ
Explanation:
In order to find the heat generated during the reaction.
q = mcΔt
The mass of the reactants in grams is determined.
The total volume of the solution is 100 mL (50 mL + 50 mL).
From this, 100 mL X 1.04g/mL gives us 104 grams of solution. (mass = Volume X Density)
The temperature variation is calculated as follows:
Δt = final temperature - initial temperature = 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 represents the heat released during the reaction between HCl and NaOH, leading to q = -5.52 × 103 J.
The reaction is exothermic.
To find the overall heat of the reaction per mole:
Using 50.0 mL of HCl, 2.00 mol HCl per 1000 mL gives us 0.100 mol HCl.
The corresponding amount of base used was also 0.100 mol of NaOH.
The energy per mole is calculated as follows:
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
Thus, the enthalpy change for the neutralization of HCl and NaOH can be expressed as ΔH = -55.177 kJ/mol
Heat absorbed by the calorimeter is calculated as: −57.32kJ − 55.177 kJ = −2.1428KJ
The heat capacity of the calorimeter is thus = −2.1428KJ÷13.5°C
= −0.1587KJ/°C
(B) Regarding the ZnCl reaction:
Determine the heat generated during the reaction.
q = mcΔt
Calculate the mass of the 100 mL HCl solution.
100 mL X 1.015g/mL equals 101.5 grams of solution. (mass = Volume X Density)
Assess the temperature change:
Δt = final temperature - initial temperature = 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 represents the heat lost in the reaction between HCl and NaOH, so q = -1.483 × 103 J.
This process is exothermic.
To determine the total heat or heat per mole for the reaction:
100.0 mL of HCl X 1.00 mol HCl /(1000 mL HCl ) results in 0.100 mol HCl.
The energy per mole is calculated as:
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
