Question

consider the following intermediate reactions. the overall chemical reaction is as follows. ch4(g) + 2o2(g) → co2(g) + 2h2o(l) δh1 = -802 kj 2h2o(g) → 2h2o(l) δh2 = -88 kj ch4(g) + 2o2(g) → co2(g) + 2h2o(g) δh3 = -890 kj what is the correct enthalpy diagram using the hess law for this system?

Explanation:

Step1: Recall Hess's law

Hess's law states that the enthalpy change of an overall reaction is the sum of the enthalpy changes of the individual steps.

Step2: Identify the overall reaction and steps

The overall reaction is $CH_4(g)+2O_2(g)
ightarrow CO_2(g) + 2H_2O(l)$ with $\Delta H_3$. The two - step reactions are $CH_4(g)+2O_2(g)
ightarrow CO_2(g)+2H_2O(g)$ with $\Delta H_1=- 802kJ$ and $2H_2O(g)
ightarrow 2H_2O(l)$ with $\Delta H_2=-88kJ$.

Step3: Calculate the overall enthalpy change

According to Hess's law, $\Delta H_3=\Delta H_1+\Delta H_2$.
Substitute the values: $\Delta H_3=-802kJ+( - 88kJ)$.
$\Delta H_3=-890kJ$.

Answer:

The overall enthalpy change for the reaction $CH_4(g)+2O_2(g)
ightarrow CO_2(g)+2H_2O(l)$ is $- 890kJ$. This value is consistent with the given $\Delta H_3$ value in the problem - statement, which validates our use of Hess's law for calculating the enthalpy change of the overall reaction.