Calculate ΔH° for each oxidation reaction. Each equation is balanced as written; remember to take into account the coefficients in determining the number of bonds broken or formed.

[ ΔHfor O2=497kJ/mol ; ΔHfor one C=O in CO2=535kJ/mol]

a.role="math" localid="1648191068323" CH4+2O2CO2+2H2O

b.2CH4+7O24CO2+6H2O


Short Answer

Expert verified

Answer

a. ΔH for the reaction is width="98">-328kJ/mol.

b. ΔH for the reaction is localid="1648191467506" -1121kJ/mol.

Step by step solution

01

Step-by-Step SolutionStep 1: Enthalpy of a reaction

The enthalpy of a reaction is determined by taking the difference of the enthalpy of reactants (sum of enthalpies of a bond broken) and the enthalpy of products (sum of enthalpies of a bond formed).

02

Formula used to calculate the enthalpy of a reaction

The formula used to calculate the enthalpy of a given reaction is:

ΔHoverall=ΔHbondsbroken-ΔHbondsformed

03

Calculating enthalpy of a reaction

a. CH4+2O2CO2+2H2O

Bonds broken:

role="math" localid="1648192060232" CH3-HΔH=4×435kJ/moland role="math" localid="1648192297307" O-OΔH=2×497kJ/mol.

The bonds formed are:

role="math" localid="1648192315890" OC-OΔH=2×535kJ/moland role="math" localid="1648192388083" HO-HΔH=4×498kJ/mol.

role="math" localid="1648193161702" ΔHoverall=ΔHbondsbroken-ΔHbondsformed=4x435+2×497kJ/mol-2×535+4×498kJ/mol=2784kJ/mol-3062kJ/mol=-328kJ/mol

b. 2CH4+7O24CO2+6H2O

Bonds broken:

CH3CH2-H(ΔH=12×410kJ/mol),O-O((ΔH=7×497kJ/mol)andC-C(ΔH=2×368kJ/mol)

Bonds formed:

OC-O(ΔH=8×535kJ/mol)andO-O((ΔH=12×498kJ/mol)

ΔHoverall=ΔHbondsbroken-ΔHbondsformed=12×410+2×368kJ/mol-8×535+12×498kJ/mol=9135kJ/mol-10256kJ/mol=-1121kJ/mol

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Most popular questions from this chapter

Calculate ΔH° for each reaction.

  1. HO·+CH4.CH3+H2O
  2. CH3OH+HBrCH3Br+H2O

For a reaction with Keq=0.8and Ea=80kJ/mol, decide which of the following statements is (are) true. Correct any false statement to make it true. Ignore entropy considerations. (a) The reaction is faster than a reaction with Keq=8and Ea=80kJ/mol. (b) The reaction is faster than a reaction with Keq=0.8and Ea=40kJ/mol. (c) G° for the reaction is a positive value. (d) The starting materials are lower in energy than the products of the reaction. (e) The reaction is exothermic.

The conversion of (CH3)3Clto (CH3)2C=CH2can occur by either a one-step or a two-step mechanism, as shown in Equations [1] and [2].

a. What rate equation would be observed for the mechanism in Equation [1]?

b. What rate equation would be observed for the mechanism in Equation [2]?

c. What is the order of each rate equation (i.e., first, second, and so forth)?

d. How can these rate equations be used to show which mechanism is the right one for this reaction?

e. Assume Equation [1] represents an endothermic reaction and draw an energy diagram for the reaction. Label the axes, reactants, products, Ea, and H°. Draw the structure for the transition state.

f. Assume Equation [2] represents an endothermic reaction and that the product of the rate-determining step is higher in energy than the reactants or products. Draw an energy diagram for this two-step reaction. Label the axes, reactants and products for each step, and the Eaand H°for each step. Label H°overall. Draw the structure for both transition states.

Use the values in Table 6.2 to calculatefor each reaction. Classify each reaction as endothermic or exothermic.

a.

b.

For a reaction with ΔH°=40kJ/mol, decide which of the following statements is (are) true. Correct any false statement to make it true. (a) ΔG°The reaction is exothermic; (b) for the reaction is positive; (c) Keq is greater than 1; (d) the bonds in the starting materials are stronger than the bonds in the product; and (e) the product is favored at equilibrium.

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