Calculate the heat of the following reaction:2CH4 + 3Cl2 → 2CHCl3 + 3 H2Given the following heats of formation:CH4 ΔH°f= -74.8 kJ/molCl2 ΔH°f= 0.0 kJ/molCHCl3 ΔH°f= -134.1 kJ/molH2 ΔH°f= 0.0 kJ/molGroup of answer choices-818.6 kJ/mol-168.6 kJ/mol-118.6 kJ/mol-618.6 kJ/mol
Question
Calculate the heat of the following reaction:2CH4 + 3Cl2 → 2CHCl3 + 3 H2Given the following heats of formation:CH4 ΔH°f= -74.8 kJ/molCl2 ΔH°f= 0.0 kJ/molCHCl3 ΔH°f= -134.1 kJ/molH2 ΔH°f= 0.0 kJ/molGroup of answer choices-818.6 kJ/mol-168.6 kJ/mol-118.6 kJ/mol-618.6 kJ/mol
Solution
To calculate the heat of the reaction, we use the formula:
ΔH°rxn = Σ ΔH°f (products) - Σ ΔH°f (reactants)
For the products, we have 2 moles of CHCl3 and 3 moles of H2. For the reactants, we have 2 moles of CH4 and 3 moles of Cl2.
So,
ΔH°rxn = [2(-134.1 kJ/mol) + 3(0.0 kJ/mol)] - [2(-74.8 kJ/mol) + 3(0.0 kJ/mol)] ΔH°rxn = [-268.2 kJ + 0.0 kJ] - [-149.6 kJ + 0.0 kJ] ΔH°rxn = -268.2 kJ + 149.6 kJ ΔH°rxn = -118.6 kJ
So, the heat of the reaction is -118.6 kJ/mol.
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