Question

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

Answer 1

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

Answer 2

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) $H_2O\rightarrow H_2+\frac{1}{2}O_2$    $\Delta H_2=71.2KJ/mole$

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

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,

$\Delta H_{formation}=[2\times \Delta H_3]+[2\times \Delta H_1]-[4\times \Delta H_2]$

where,

n = number of moles

$\Delta H_{formation}=[2\times (283)]+[2\times (94.5)]-[4\times (71.2)]=470.2KJ/mole$

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