Question

Calculate ΔHrxn for the following reaction: CH4(g)+2O2(g)→CO2(g)+2H2O(l) Use the following reactions and given ΔH′s. CH4(g)+O2(g)→CH2O(g)+H2O(g), ΔH = -284 kJ CH2O(g)+O2(g)→CO2(g)+H2O(g), ΔH = -527 kJ H2O(l)→H2O(g), ΔH = 44.0 kJ

Answer 1

Answer:

ΔH of CH4(g)+2O2(g)→CO2(g)+2H2O(l) is -899 KJ

Explanation:

step 1: write down thye reaction and rearrange them in a way that similar contents cancel out to produce the required reaction and add the enthalpy change of each.

CH4(g)+O2(g)→CH2O(g)+H2O(g)   ΔH = -284 KJ

CH2O(g)+O2(g)→CO2(g)+H2O(g)  ΔH = -527 KJ

2H2O(g)→2H2O(l)    ΔH = 44 KJ x -2

the minus is because the last reaction has been reversed and multiplied by 2 so that the gaseous state H2O cancels with the others to leave liquid H2O in the required final reaction.

ΔHrxn = (-284)+(-527)+[44 x (-2)] = -899 KJ

Answer 2

ΔHrxn for the following reaction: CH4(g)+2O2(g)→CO2(g)+2H2O(l) : ΔH = -899 kJ

Further explanation

The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation

The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔH °)

Determination of the enthalpy of formation of a reaction can be through a calorimetric experiment, based on the principle of Hess's Law, enthalpy of formation table, or from bond energy data

Delta H reaction (ΔH) is the amount of heat / heat change between the system and its environment

(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)

The value of ° H ° can be calculated from the change in enthalpy of standard formation:

∆H ° rxn = ∑n ∆H ° (product) - ∑n ∆H° (reactants)

ΔH∘rxn = ΔH∘ the product (s) if ∆H ° (reactants) = 0

For the corresponding reaction we use stages of several reactions to get the desired reaction ΔH

CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l)

Use the following reactions and given ΔH′s.

1. CH4 (g) + O2 (g) → CH2O (g) + H2O (g), ΔH = -284 kJ

2. CH2O (g) + O2 (g) → CO2 (g) + H2O (g), ΔH = -527 kJ

3. H2O (l) → H2O (g), ΔH = 44.0 kJ

Of the three ΔH known reactions we can combine to get ΔH unknown reactions

We add up the three reactions by first reversing and multiplying reaction 3 by the coefficient 2

1. CH4 (g) + O2 (g) → CH2O (g) + H2O (g), ΔH = -284 kJ

2. CH2O (g) + O2 (g) → CO2 (g) + H2O (g), ΔH = -527 kJ

3. 2H2O (g) → 2H2O (l), ΔH = -88.0 kJ

 We remove CH2O (g) because it is left and right in the reaction so the sum of the three reactions becomes:

CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l) ΔH = -899 kJ

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