Question

The mea sured enthalpy change for burning ketene (CH2CO) CH2CO(g) 1 2 O2(g)88n 2 CO2(g) 1 H2O(g) is DH1 5 2981.1 kJ at 25°C. The enthalpy change for burning methane CH4(g) 1 2 O2(g)88n CO2(g) 1 2 H2O(g) is DH2 5 2802.3 kJ at 25°C. Calculate the enthalpy change at 25°C for the reaction 2 CH4(g) 1 2 O2(g)88n CH2CO(g) 1 3 H2O(g)

Answer 1

Answer : The value of enthalpy change for the reaction is, 2623.5 kJ

Explanation :

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 given main reaction is,

$2CH_4(g)+2O_2(g)\rightarrow CH_2CO(g)$    $\Delta H=?$

The intermediate balanced chemical reaction will be,

(1) $CH_2CO(g)+2O_2(g)\rightarrow CO_2(g)+H_2O(g)$     $\Delta H_1=2981.1kJ$

(2) $CH_4(g)+2O_2(g)\rightarrow CO_2(g)+H_2O$    $\Delta H_2=2802.3kJ$

Now we will reverse the reaction 1, multiplying reaction 2 by 2 and then adding all the equations, we get :

(1) $CO_2(g)+H_2O(g)\rightarrow CH_2CO(g)+2O_2(g)$     $\Delta H_1=-2981.1kJ$

(2) $2CH_4(g)+4O_2(g)\rightarrow 2CO_2(g)+2H_2O$    $\Delta H_2=2\times 2802.3kJ=5604.6kJ$

The expression for enthalpy of change will be,

$\Delta H=\Delta H_1+\Delta H_2$

$\Delta H=(-2981.1kJ)+(5604.6kJ)$

$\Delta H=2623.5kJ$

Therefore, the value of enthalpy change for the reaction is, 2623.5 kJ