Question

Given the following reactions and subsequent ∆H values what is the ∆H for the reaction below?

4CO2(g) + 2H2O(g) 2C2H2(g) + 5O2(g)
Given:
C2H2+2H2----> C2H6 ∆H = -94.5kJ
H2O ----> H2 + 1/2 O2 ∆H = 71.2kJ
C2H6 + 7/2 O2 -----> 2CO2 + 3H2O ∆H = -283kJ

Answer 1

Answer : The enthalpy change of the reaction is, -470.2 KJ/mole

Solution :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

The balanced chemical reaction are,

(1)
`C_2H_2+H_2\rightarrow C_2H_6`
`\Delta H_1=-94.5KJ/mole`

(2)
`H_2O\rightarrow H_2+(1)/(2)O_2`
`\Delta H_2=71.2KJ/mole`

(3)
`C_2H_6+(7)/(2)O_2\rightarrow 2CO_2+3H_2O`
`\Delta H_3=-283KJ/mole`

The balanced main chemical reaction will be,

`4CO_2(g)+2H_2O(g)\rightarrow 2C_2H_2(g)_5O_2(g)`
`\Delta H=?`

First reverse the reaction 3 then adding the twice of reaction 3, twice of reaction 1 and then subtracting four times of reaction 2 from the addition of two reaction 3 and 1, we get the enthalpy change of the reaction.

The expression for enthalpy change of the reaction is,

`\Delta H_(formation)=[2* \Delta H_3]+[2* \Delta H_1]-[4* \Delta H_2]`

where,

n = number of moles

`\Delta H_(formation)=[2* (283)]+[2* (94.5)]-[4* (71.2)]=470.2KJ/mole`

Therefore, the enthalpy change of the reaction is, -470.2 KJ/mole

Answer 2

The heat of a reaction can be calculated from the heat of formation of each substance in the reaction. It is calculated as the sum of the heat of formation of each substance.

C2H2+2H2----> C2H6 ∆H = (-94.5kJ)
2(H2O ----> H2 +1/2O2) ∆H = 2(71.2kJ)
C2H6 + 7/2 O2 -----> 2CO2 + 3H2O ∆H = (-283kJ)
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C2H2 + 5/2O2 = 2CO2 + H2O ∆H = -235.1 kJ

4CO2 + 2H2O = 2C2H2 + 5O2 ∆H = 470.2 kJ