Question

Calculate ΔHrxnΔHrxn for the following reaction: CH4(g)+2O2(g)→CO2(g)+2H2O(l)CH4(g)+2O2(g)→CO2(g)+2H2O(l) Use the following reactions and given ΔHΔH values. CH4(g)+O2(g)→CH2O(g)+H2O(g)CH4(g)+O2(g)→CH2O(g)+H2O(g), ΔH=−ΔH=−284 kJkJ CH2O(g)+O2(g)→CO2(g)+H2O(g)CH2O(g)+O2(g)→CO2(g)+H2O(g), ΔH=−ΔH=−527 kJkJ H2O(l)→H2O(g)H2O(l)→H2O(g), ΔH=ΔH= 44.0 kJ

Answer 1

Answer: the enthalpy of reaction is, -155 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.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The final reaction is:

$CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)$    $\Delta H_{rxn}=?$

The intermediate balanced chemical reaction will be,

(1) $CH_4(g)+O_2(g)\rightarrow CH_2O(g)+H_2O(g)$     $\Delta H_1=-284kJ$

(2) $CH_2O(g)+O_2(g)\rightarrow CO_2(g)+H_2O(g)$    $\Delta H_2=-527kJ$

(3) $H_2O(l)\rightarrow H_2O(g)$    $\Delta H_3=44.0kJ$

Now multiplying (3) by 2 and adding all the equations, we get :

$\Delta H_{rxn}=\Delta H_1+\Delta H_2+2\times \Delta H_3$

$\Delta H_{rxn}=(-284)+(-527)+2\times (44)$

$\Delta H_{rxn}=-155kJ$

Therefore, the enthalpy of reaction is, -155 kJ