Question

The standard enthalpy of formation of methane (\(CH_4\)), using the standard enthalpy of reaction values from the following reaction steps, is \(-x\) kcal/mol. The value of \(x\) is _ _ _. (round off to one decimal place)

Hint:

Hess's law states that the total enthalpy change for a reaction is independent of the pathway and depends only on initial and final states.

Answer 1

Correct Answer: 17.8

Step 1: List the given steps.
We are given $\Delta H_f(H_2O) = -68.3$, $\Delta H_f(CO) = -26.4$ kcal/mol, and the step $CO + 3H_2 \rightarrow CH_4 + H_2O$ with $\Delta H = -59.7$ kcal/mol.

Step 2: Target reaction.
We want the formation of methane, $C + 2H_2 \rightarrow CH_4$.

Step 3: Write the Hess relation.
For the given step, \[ \Delta H = \Delta H_f(CH_4) + \Delta H_f(H_2O) - \Delta H_f(CO) \]

Step 4: Substitute and solve.
\[ -59.7 = \Delta H_f(CH_4) + (-68.3) - (-26.4) = \Delta H_f(CH_4) - 41.9 \] so $\Delta H_f(CH_4) = -59.7 + 41.9 = -17.8$ kcal/mol, giving $x = 17.8$.

Step 5: Answer.
\[ \boxed{17.8\ \text{kcal/mol}} \]