Answer to Question #203847 in General Chemistry for Monique Keith Anto

Question

Topic: Enthalpy and Hess Law

1. Calculate the heat of hydrogenation of ethane, C2H4 given the following thermochemical equations:

2 C (graphite) + 3 H2 (g) ---> C2H6 (g). ∆Hf= -84.5 kJ/mol

2 C (graphite) + 2 H2 (g) ---> C2H4 (g). ∆Hf= 52.3 kJ/mol

2. Calculate the ∆H for the reaction:

CS2 (I) + 2 O2 (g) CO2 (g) + 2 SO2 (g)

Given:

∆Hf CO2 (g) = - 393.5 kJ/mol

∆Hf SO2 = - 296.8 kJ/mol

∆Hf CS2 (I) = 87.9 kJ/mol

Answer 1

1) we have to calculate heat of hydrogenation of ethane $C_2H_6$

2 C (graphite) + 3 H2 (g) ---> C2H6 (g). ∆Hf= -84.5 kJ/mol..........................1

2 C (graphite) + 2 H2 (g) ---> C2H4 (g). ∆Hf= 52.3 kJ/mol.............................2

for that we will subtract equation 2 from equation 1.

and we will get,

$C_ 2 ​ H _ 4 ​ (g)+H_ 2 ​ (g)⟶C _ 2 ​ H _ 6 ​ (g)$

and $\Delta H_f =\Delta H_{f1}-(-\Delta H_{f2})$

$\Delta H_{f}=-84.5+52.3\\$

$\Delta H_{f}(ethane)= 32.2$ kJ/mol

2) ∆H for the reaction:

CS₂ (I) + 2 O₂ (g ) -----> CO₂ (g) + 2 SO₂ (g)

we know that,

$\Delta{H}_{reaction}=\sum{\Delta{H_{products}-\sum\Delta{H}_{reactants}}}$

$=−393.5+2(−296.8)−87.9$

$=−1075kJ/mol$