Answer: The enthalpy change of the reaction comes out to be -890 kJ
Explanation:
To calculate the enthalpy of the reaction divided in multiple steps, we use Hess's law. This law states that the heat absorbed or evolved in a given chemical equation remains same whether the process occurs in one step or several steps.
To apply this law, the enthalpy of the chemical equation is treated as ordinary algebraic expression and it can be added or subtracted to form the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes that occur during intermediate reactions.
For the given overall reaction:
[tex]CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)[/tex]
The two intermediate reactions are as follows:
Reaction 1: [tex]CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g);\Delta H_1=-802kJ[/tex]
Reaction 2: [tex]H_2O(g)\rightarrow 2H_2O(l);\Delta H_2=-88kJ[/tex]
By adding both the reactions, we will get the overall enthalpy of the reaction, which is:
[tex]\Delta H_{rxn}=\Delta H_1+\Delta H_2\\\\\Delta H_{rxn}=[(-802)+(-88)]kJ=-890kJ[/tex]
Hence, the enthalpy change of the reaction comes out to be -890 kJ
Answer : The value of [tex]\Delta H_{rxn}[/tex] for the overall reaction is, -890 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.
For the given overall reaction:
[tex]CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)[/tex]
The two intermediate reactions are as follows:
(1) [tex]CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g);\Delta H_1=-802kJ[/tex]
(2) [tex]2H_2O(g)\rightarrow 2H_2O(l);\Delta H_2=-88kJ[/tex]
Now adding both the reactions, we will get the overall enthalpy of the reaction, which is:
[tex]\Delta H_{rxn}=\Delta H_1+\Delta H_2\\\\\Delta H_{rxn}=[(-802)+(-88)]kJ=-890kJ[/tex]
Hence, the value of [tex]\Delta H_{rxn}[/tex] for the overall reaction is, -890 kJ
