Respuesta :
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
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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