First
Question: Calculate the heat of hydrogenation of ethane, C2H4 given the following thermochemical equations: 2 C(graphite) + 3 H2 (g) —> C2H6 (g) ΔHf= - 84.5 kJ/mol 2 C (graphite) + 2 H2 (g) —> C2H4 (g) ΔHf= 52.3 kJ/mol.
Answer:
Step 1
Step 2
Second
Question:
Answer:
Step 1
Given data contains,
ΔHfCO2 (g) = − 393.5 kJ/mol; ΔHfSO2 (g)= −296.8 kJ/mol; ΔHfCS2 (l) = 87.9kJ/molΔHfCO2 (g) = - 393.5 kJ/mol; ΔHfSO2 g= -296.8 kJ/mol; ΔHfCS2 (l) = 87.9kJ/mol
The given chemical reaction is,
CS2(l)+2O2(g)−→CO2(g)+2SO2(g)CS2l+2O2g→CO2g+2SO2g
Step 2
The ∆Hreaction∆Hreaction is calculated as,
∆Hreaction=∑∆Hproducts−∑∆Hreactants∆Hreaction=[−393.5 kJ/mol+2(−296.8 kJ/mol)]−[87.9 kJ/mol]∆Hreaction=−1075 kJ/mol∆Hreaction=∑∆Hproducts-∑∆Hreactants∆Hreaction=-393.5 kJ/mol+2-296.8 kJ/mol-87.9 kJ/mol∆Hreaction=-1075 kJ/mol
The ∆Hreaction∆Hreaction is −1075 kJ/mol-1075 kJ/mol.
Third
Question:
Calculate the ΔH for the reaction: CS2 (l) + 2 O2 (g) —> CO2 (g) + 2 SO2 (g)
Given: ΔHf o CO2 (g) = - 393.5 kJ/mol; ΔHf o SO2 = -296.8 kJ/mol; ΔHf o CS2 (l) = 87.9kJ/mol
Answer:
Step 1
Heat of reaction or enthalpy change (∆H∘rxn)(∆H∘rxn)∆Hrxn∘ is defined as heat released and absorbed during a reaction. It is calculated by the difference of enthalpies of formation of products and enthalpies of formation of reactants.
Note: The given reaction is not balanced but the correct balanced reaction is used in the calculation of enthalpy of reaction.