Chemistry 111: Enthalpy change

Question

1.      Consider the following reaction:

2 CH₄(g) + O₂(g)   à   2CH₃OH(g) DH=-252.8 kJ

1. 
   
2. Is this reaction exothermic or endothermic?

1. Calculate the amount of heat transferred when 35.0 g of CH₄(g) is used up by this reaction at constant pressure

1. For a given sample of CH₄ the enthalpy change during the reaction is -105.1 kJ. How many grams of CH₃OH, are produced?

Answer 1

1.      Consider the following reaction:

2 CH₄(g) + O₂(g)   à   2CH₃OH(g) DH=-252.8 kJ

a) This reaction is exothermic because delta H is negative.

Endothermic reactions occur when delta H is positive. Heat is considered a reactant in the equation. Exothermic reactions occur when delta H is negative. Heat is considered a product in the equation.

b) The delta H given with the equation is the amount of heat released per one mole of the equation:

This means H = -252.8 kJ / 2 mole CH4

H = -252.8 kJ / 1 mole O2

H = -252.8 kJ / 2 mole CH3OH

Viewing delta H in this way gives insight to how the quantity of CH4 used could affect the energy released, because now there is a direct relationship between the moles of CH4 and the energy of the reaction.

Path

35.0 g CH4 --> moles CH4 --> energy released

Using the delta H expression along with the molar mass of CH4 we can use stoichiometry to get the result. Remember, the fractional expressions for energy and molar mass can be used in their original form or flipped because the relationship remains true. The diagonals of the stoichiometric ratios should match their units so those units will disappear.

Molar Mass:

C: 1 x 12.01g

H: 4 x 1.01 g

MM = 16.05 g / 1 mole CH4

35.0 g CH4 x 1 mol CH4 / 16.05 g CH4 x -252.8 kJ / 2 mol CH4 = -275.6 kJ released

c) Part c will follow exactly the same steps as part b. However, this time the question asks for the amount of CH3OH produced. Using the delta H expression for CH3OH and molar mass of CH3OH we will complete a similar stoichiometric calculation, this time converting from units of energy back to units of grams.

Path -105.1 kJ --> moles CH3OH --> grams CH3OH

Molar Mass:

C: 1 x 12.01g

H: 4 x 1.01g

O: 1 x 16.00g

MM = 32.05 g / 1 mol CH3OH

delta H = -252.8 kJ / 2 mol CH3OH

-105.1 kJ x 2 mol CH3OH / -252.8 kJ x 32.05 g CH3OH / 1 mol CH3OH = 26.6 g CH3OH