Question

Calculate the standard enthalpy change of following reaction:
$CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_{2}O_{(l)}$
if $\Delta_{f}H^{\circ}(CH_{4}) = -75~kJ~mol^{-1}$, $\Delta_{f}H^{\circ}(CO_{2}) = -390~kJ~mol^{-1}$, $\Delta_{f}H^{\circ}(H_{2}O) = -286~kJ~mol^{-1}$

• A. -887.00 kJ
• B. -1325.00 kJ
• C. -1035.00 kJ
• D. -1770.00 kJ

Hint:

Always remember that standard enthalpy of formation for elements in their natural state is zero.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The enthalpy change of a reaction (\( \Delta_{\text{r}}\text{H}^{\circ} \)) can be calculated from the standard enthalpies of formation of the products and reactants.

Step 2: Key Formula or Approach:
\[ \Delta_{\text{r}}\text{H}^{\circ} = \sum \Delta_{\text{f}}\text{H}^{\circ} (\text{products}) - \sum \Delta_{\text{f}}\text{H}^{\circ} (\text{reactants}) \]

Step 3: Detailed Explanation:
The reaction is: \( \text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \)
Note: \( \Delta_{\text{f}}\text{H}^{\circ} \) for an element in its standard state (\( \text{O}_2 \)) is zero.
\[ \Delta_{\text{r}}\text{H}^{\circ} = [ \Delta_{\text{f}}\text{H}^{\circ}(\text{CO}_2) + 2 \times \Delta_{\text{f}}\text{H}^{\circ}(\text{H}_2\text{O}) ] - [ \Delta_{\text{f}}\text{H}^{\circ}(\text{CH}_4) + 2 \times \Delta_{\text{f}}\text{H}^{\circ}(\text{O}_2) ] \]
Substituting the values:
\[ \Delta_{\text{r}}\text{H}^{\circ} = [ (-390) + 2 \times (-286) ] - [ (-75) + 0 ] \]
\[ \Delta_{\text{r}}\text{H}^{\circ} = [ -390 - 572 ] - [ -75 ] \]
\[ \Delta_{\text{r}}\text{H}^{\circ} = [ -962 ] + 75 \]
\[ \Delta_{\text{r}}\text{H}^{\circ} = -887 \text{ kJ} \]

Step 4: Final Answer:
The standard enthalpy change of the reaction is \( -887.00 \text{ kJ} \).