Question

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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

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

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

(a) Table of enthalpies of formation of reactants and products $\begin{array}{cc}\textbf{Substance} & \textbf{ \Delta_{\text{f}} H/(kJ/mol}) \\\text{C _{3} H _{8} (g)} & -103.85 \\\text{O}_{2}\text{(g)} & 0 \\\text{CO}_{2}\text{(g)} & -393.51 \\\text{H _{2} O(g)} & -241.82\\\end{array}$

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(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.

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Answer 2

Answer:ddsss

Explanation:saawd