Question

90 POINTSSSS!!!

Step 3: Determine the amount of energy change in the reaction. a) Use the table of enthalpy values (Table A) provided in the Student Worksheet to locate the enthalpy of formation (∆Hf) for each reactant and each product. Record these values along with the reactants and products in Table B of the Student Worksheet. b) Determine the total enthalpy of the reactants and the total enthalpy of the products. Record these values in Table C of the Student Worksheet. c) Use the following formula to find the net change in enthalpy for the reaction and to determine whether the reaction is endothermic or exothermic.

Answer 1

See explanation

Step-by-step explanation:

The reaction equation is;

C3H8 (g) + 5O2(g) -------> 4H2O(g) + 3CO2(g)

From the formula;

Total enthalpy of reactants = (ΔHf of Reactant 1 x Coefficient) + (ΔHf of Reactant 2 x Coefficient)

Total enthalpy of products= (ΔHf of Product 1 x Coefficient) + (ΔHf of Product 2 x Coefficient)

Hence;

Total enthalpy of reactants =[(-103.85 * 1) + (0 * 5)] = -103.85 + 0 = -103.85 KJ/mol

Total enthalpy of products= [(-393.51 * 4) +(-241.82 * 3)] = (-1574.04) + (-483.64) = -2057.68 KJ/mol

Step-by-step explanation:

Answer 2

`\large \boxed{\text{-2043.96 kJ/mol}}`

Step-by-step explanation:

Assume the reaction is the combustion of propane.

Word equation: propane plus oxygen produces carbon dioxide and water

Chemical eqn: C₃H₈(g) + O₂(g) ⟶ CO₂(g) + H₂O(g)

Balanced eqn: C₃H₈(g) + 5O₂(g) ⟶ 3CO₂(g) + 4H₂O(g)

(a) Table of enthalpies of formation of reactants and products

`\begin{array}{cc}\textbf{Substance} & \textbf{$\Delta_{\text{f}}$H/(kJ/mol}) \\\text{C$_(3)$H$_(8)$(g)} & -103.85 \\\text{O}_(2)\text{(g)} & 0 \\\text{CO}_(2)\text{(g)} & -393.51 \\\text{H$_(2)$O(g)} & -241.82\\\end{array}`

(b)Total enthalpies of reactants and products

`\Delta_{\text{rxn}}H^(\circ) = \sum \left( \Delta_{\text{f}} H^(\circ) \text{products}\right) - \sum \left (\Delta_{\text{f}}H^(\circ) \text{reactants} \right)\\= \text{-2147.81 kJ/mol - (-103.85 kJ/mol)}\\= \text{-2147.81 kJ/mol + 103.85 kJ/mol}\\= \textbf{-2043.96 kJ/mol}\\\text{The enthalpy change is $\large \boxed{\textbf{-2043.96 kJ/mol}}$}`

ΔᵣH° is negative, so the reaction is exothermic.