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Question: What is the heat of combustion for 4 g of sulfur if heat of formation of $SO_{2(g)}$ is -298 kJ $mol...

What is the heat of combustion for 4 g of sulfur if heat of formation of SO2(g)SO_{2(g)} is -298 kJ mol−1mol^{-1}?

A

18.6 kJ

B

-29.8 kJ

C

-37.25 kJ

D

10.6 kJ

Answer

-37.25 kJ

Explanation

Solution

For the combustion reaction

S(s)+O2(g)→SO2(g)S_{(s)} + O_{2(g)} \rightarrow SO_{2(g)}

the enthalpy change is given by the heat of formation of SO2SO_2, i.e., −298 kJ/mol-298\,\text{kJ/mol}.

  1. Calculate moles of sulfur in 4 g:

Moles of S=4 g32 g/mol=0.125 mol\text{Moles of S} = \frac{4\,\text{g}}{32\,\text{g/mol}} = 0.125\,\text{mol}

  1. Calculate heat released:

Heat released=0.125×(−298 kJ/mol)=−37.25 kJ\text{Heat released} = 0.125 \times (-298\,\text{kJ/mol}) = -37.25\,\text{kJ}