Answer to Question #334504 in General Chemistry for peya

Question

1. Calculate the ΔH for the reaction: 

CS2 (l) + 2 O2 (g)  �  CO2 (g) + 2 SO2 (g)

Given: 

ΔHfCO2 (g) = - 393.5 kJ/mol; 

ΔHfSO2 = -296.8 kJ/mol; 

ΔHfCS2 (l) = 87.9kJ/mol

 

Answer 1

The enthalpy of reaction is the difference of sum of the enthalpies of formation of the products with the sum of the enthalpies of formation of reactants:

$\Delta H = \sum \Delta H_f^{prod} - \sum \Delta H_f^{react}$

Note that the enthalpy of formation of oxygen equals to 0 ($\Delta H_f (O_2) = 0 kJ/mol$ )

$\Delta H =(\Delta H_f(CO_2) +2\Delta H_f(SO_2)) - (\Delta H_f(CS_2) + 2\Delta H_f(CS_2))$

$\Delta H= (-393.5 +2\times (-296.8)) - (87.9+2 \times 0) = -1075 kJ/mol$