![Using Hess's Law, calculate the heat of formation of propane ( extC3 extH8) given the following reactions and their enthalpy changes: ... [list of reactions]](/_next/image?url=https%3A%2F%2Fohijasnjvprzodcmxltt.supabase.co%2Fstorage%2Fv1%2Fobject%2Fpublic%2Fwhatsapp-images%2F1777862720325-aee8e65155d06d1d.png&w=3840&q=75)
This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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You're on a roll — Here's how to calculate the heat of formation of propane using Hess's Law: State Hess's Law: Hess's Law states that the total enthalpy change for a chemical reaction is independent of the pathway taken, as long as the initial and final conditions are the same. This means that if a reaction can be expressed as a sum of several steps, the enthalpy change for the overall reaction is the sum of the enthalpy changes for each step. Calculate the heat of formation of propane: Step 1: Write the target equation for the formation of propane (C_3H_8). 3C(s) + 4H_2(g) C_3H_8(g) Step 2: List the given reactions and their enthalpy changes. 1. C(s) + O_2(g) CO_2(g); H_1 = -393 kJ/mol 2. H_2(g) + (1)/(2)O_2(g) H_2O(l); H_2 = -286 kJ/mol 3. C_3H_8(g) + 5O_2(g) 3CO_2(g) + 4H_2O(l); H_3 = -2209 kJ/mol Step 3: Manipulate the given reactions to match the target equation. • To get 3C(s) on the reactant side, multiply Reaction 1 by 3: 3C(s) + 3O_2(g) 3CO_2(g); H_1' = 3 × (-393 kJ/mol) = -1179 kJ/mol • To get 4H_2(g) on the reactant side, multiply Reaction 2 by 4: 4H_2(g) + 2O_2(g) 4H_2O(l); H_2' = 4 × (-286 kJ/mol) = -1144 kJ/mol • To get C_3H_8(g) on the product side, reverse Reaction 3. This changes the sign of its enthalpy change: 3CO_2(g) + 4H_2O(l) C_3H_8(g) + 5O_2(g); H_3' = -(-2209 kJ/mol) = +2209 kJ/mol Step 4: Add the manipulated reactions and their enthalpy changes. Adding the three manipulated reactions: 3C(s) + 3O_2(g) & 3CO_2(g) \\ 4H_2(g) + 2O_2(g) & 4H_2O(l) \\ 3CO_2(g) + 4H_2O(l) & C_3H_8(g) + 5O_2(g) \\ 3C(s) + 4H_2(g) + (3+2)O_2(g) + 3CO_2(g) + 4H_2O(l) & 3CO_2(g) + 4H_2O(l) + C_3H_8(g) + 5O_2(g) Cancel out common species (5O_2(g), 3CO_2(g), 4H_2O(l)) from both sides to get the target equation: 3C(s) + 4H_2(g) C_3H_8(g) Step 5: Calculate the total enthalpy change. H_f(C_3H_8) = H_1' + H_2' + H_3' H_f(C_3H_8) = (-1179 kJ/mol) + (-1144 kJ/mol) + (+2209 kJ/mol) H_f(C_3H_8) = -2323 kJ/mol + 2209 kJ/mol H_f(C_3H_8) = -114 kJ/mol The heat of formation of propane is -114 kJ/mol. What's next?