Question

2FeS2(g)+1/2O2(g)−→Fe2O3+4SO2(g). TakingtheenthalpyofformationofFeS2(g),Fe2O3 andSO2(g) to be -180, -824 and -297 kJmol−1. If the starting material FeS2 is 0.1, what is the value of enthalpy of reaction and also if the C −v of the calorimeter is 10 kJK−1, what will be the change in the temperature.

Answer 1

2FeS₂ (g) + 1/2 O₂ $\to$ Fe₂O₃ + 4 SO₂ (g)

-180 _____0__________ - 824____ - 297

$\Delta H$ = - 824 - 4(-297) - 2(-180)

= -824 -1188 +360 KJ/mol

= - 1652 KJ/mol

$\Delta H$ = - n c $\Delta T$

- 1652 = - 0.1 × 10 × $\Delta T$

$\Delta T$ = 165.2