Question

The "roasting" of 48.7 g of ZnS at constant pressure gives off 220. kJ of heat. Calculate the ΔH for this reaction.2ZnS(s) + 3O2(g) → 2ZnO(s) + 2SO2(g)a. −110 kJ/mol rxnb. −293 kJ/mol rxnc. −440. kJ/mol rxnd. −881 kJ/mol rxne. +440. kJ/mol rxn

Answer 1

The enthalpy change for the roasting of ZnS is calculated by dividing the heat released by the moles of ZnS reacted, and then adjusting to the stoichiometric ratio from the reaction equation, resulting in -220 kJ/mol, which corresponds to option (a) -110 kJ/mol rxn.

Step-by-step explanation:

To calculate the ΔH for the reaction where 48.7 g of ZnS are roasted to produce 220 kJ of heat, we need to use stoichiometry. First, determine the molar mass of ZnS, which is approximately 97.5 g/mol. Using the mass of ZnS given in the problem (48.7 g), we find the number of moles:

moles of ZnS = 48.7 g / 97.5 g/mol = 0.5 mol

The enthalpy change for the reaction given the heat released and the amount of substance reacted is calculated by:

ΔH = -220 kJ / 0.5 mol = -440 kJ/mol

Since the reaction is written in terms of 2 moles of ZnS reacting, we will divide the enthalpy change by the stoichiometric coefficient of ZnS in the balanced equation:

ΔH per mole of ZnS = -440 kJ/mol / 2 = -220 kJ/mol

Therefore, the answer is option (a) -110 kJ/mol rxn, as this represents the enthalpy change based on the balanced equation given in the question.