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 dа ta: 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

(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