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1) Calculate the heat transferred to the system when 1.00 mol of a perfect gas e

ID: 717181 • Letter: 1

Question

1) Calculate the heat transferred to the system when 1.00 mol of a perfect gas expands reversibly at a constant temperature of 25°C so that its volume doubles. 2) The constant pressure molar heat capacity of zinc is 25.40 J K-1 mo-1 at 298 K Calculate the constant pressure specific heat capacity of zinc at this temperature. 3) Use the following data to determine the standard enthalpy of reaction at 298 K for the addition of hydrogen chloride, HCI, to ethene, C2H4 C2H4(g) + HCI(g) C2H5CI(g) DfHo (298 K) /kJ mol-1 C2H4 +52.2 HCI-92.3 C2H5CI-109.8 4) When 7.82 g of benzaldehyde, C6H5CHO, was burned in a bomb calorimeter at 298.15 K, the heat released was 259.2 kJ. Calculate the enthalpy of combustion of benzaldehyde at this temperature.

Explanation / Answer

Solution :-

Q1)Moles of gas n = 1.00 mol

Temperature T= 25 C +273 = 298 K

Energy tranffered = ?

Gas expands until volumes doubles therefore ratio of the V2/V1 = 2

q = nRT ln(V2/V1)

q= 1.00 mol * 8.314 J/mol K * 298 K * ln(2)

q= 1717 J

Therefore energy transferred is 1717 J

Q2) molar heat capacity of zinc = 25.40 J/mol K

Specific heat capacity of zinc = ?

Specific heat capacity of zinc = molar heat capacity / molar mass of Zinc

                                                    = 25.40J/mol K / 65.38 g /mol

                                                     = 0.3885 J/gK

Therefore specific heat capacity of zinc is 0.3885 J/gK

Q3) C2H4(g) + HCl(g) ---- > C2H5Cl (g)

Delta H rxn = sum of Delta Hf product – sum of delta Hf reactant

                     = [1*C2H5Cl] –[(1*C2H4)+(1*HCl)]

                     =[-109.8*1] –[(52.2*1)+(-92.3*1)]

                     = -69.7 kJ/mol

Therefore the enthalpy change of reaction is -69.7 kJ/mol

Q4) 7.82 g benzaldehyde gives 259.2 kJ heat

Enthalpy of combustion of benzaldehyde = ?

Molar mass of benzaldehyde = 106.121g /mol

Lets calculate the heat given by per mole benzaldehyde

259.2 kJ * 106.121 g per mol / 7.82 g = 3517 kJ/mol

Since heat is given out therefore it exothermic process hence we have to use the negative sign to the enthalpy of combustion value

Therefore enthalpy of combustion of benzaldehyde is -3517 kJ/mol