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

(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

a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.