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: Write the given reactions and enthalpy values.
\[ H_2O \rightarrow H_2+\frac{1}{2}O_2 \qquad \Delta H=+68.3\ \text{kcal/mol} \] \[ C(graphite)+\frac{1}{2}O_2 \rightarrow CO \qquad \Delta H=-26.4\ \text{kcal/mol} \] \[ CO+3H_2 \rightarrow CH_4+H_2O \qquad \Delta H=-59.7\ \text{kcal/mol} \]

Step 2: Identify the target reaction.
We need the standard enthalpy of formation of methane:
\[ C(graphite)+2H_2 \rightarrow CH_4 \]

Step 3: Reverse the first reaction.
To eliminate \(H_2O\), reverse the first equation:
\[ H_2+\frac{1}{2}O_2 \rightarrow H_2O \] Reversing changes the sign of enthalpy:
\[ \Delta H=-68.3\ \text{kcal/mol} \]

Step 4: Add the three equations.
Now add:
\[ C+\frac{1}{2}O_2 \rightarrow CO \] \[ CO+3H_2 \rightarrow CH_4+H_2O \] \[ H_2+\frac{1}{2}O_2 \rightarrow H_2O \quad \text{(reversed form used appropriately)} \] Instead, more systematically, use Hess law.

Step 5: Use enthalpy relation.
From the third equation:
\[ \Delta H_3 = \Delta H_f(CH_4)+\Delta H_f(H_2O)-\Delta H_f(CO) \] We know:
\[ \Delta H_f(H_2O)=-68.3\ \text{kcal/mol} \] \[ \Delta H_f(CO)=-26.4\ \text{kcal/mol} \] and
\[ \Delta H_3=-59.7\ \text{kcal/mol} \]

Step 6: Substitute values.
\[ -59.7 = \Delta H_f(CH_4)+(-68.3)-(-26.4) \] \[ -59.7 = \Delta H_f(CH_4)-68.3+26.4 \] \[ -59.7 = \Delta H_f(CH_4)-41.9 \]

Step 7: Solve for \(\Delta H_f(CH_4)\).
\[ \Delta H_f(CH_4) = -59.7+41.9 \] \[ =-17.8\ \text{kcal/mol} \]
Thus,
\[ -x=-17.8 \] Therefore,
\[ x=17.8 \]

Step 8: Final conclusion.
Hence, the value of \(x\) is
\[ \boxed{17.8} \]