Using the following thermochemical data.
`C(S)+O_(2)(g)rarr CO_(2)(g), Delta H=94.0` Kcal
`H_(2)(g)+1//2O_(2)(g)rarr H_(2)O(l), Delta H=-68.0` Kcal
`CH_(3)COOH(l)+2O_(2)(g)rarr 2O_(2)(g)rarr 2CO_(2)(g)+2H_(2)O(l), Delta H=-210.0` Kcal
The heat of formation of acetic acid is :-

Options

A. 116.0 Kcal
B. `-116.0` Kcal
C. `-114.0` Kcal
D. `+114.0` Kcal

Answer

To calculate the heat of formation of acetic acid (CH₃COOH) using the given thermochemical data, we will follow these steps:

### Step 1: Write the formation reaction for acetic acid
The formation reaction for acetic acid from its elements in their standard states is:
\[ \text{C (s)} + 2 \text{H}_2 \text{(g)} + \text{O}_2 \text{(g)} \rightarrow \text{CH}_3\text{COOH (l)} \]

### Step 2: Identify the relevant reactions and their enthalpy changes
We have the following reactions and their enthalpy changes:
1. \( \text{C (s)} + \text{O}_2 \text{(g)} \rightarrow \text{CO}_2 \text{(g)}, \Delta H = -94.0 \text{ kcal} \)
2. \( \text{H}_2 \text{(g)} + \frac{1}{2} \text{O}_2 \text{(g)} \rightarrow \text{H}_2\text{O (l)}, \Delta H = -68.0 \text{ kcal} \)
3. \( \text{CH}_3\text{COOH (l)} + 2 \text{O}_2 \text{(g)} \rightarrow 2 \text{CO}_2 \text{(g)} + 2 \text{H}_2\text{O (l)}, \Delta H = -210.0 \text{ kcal} \)

### Step 3: Reverse the third reaction
To find the heat of formation of acetic acid, we need to reverse the third reaction:
\[ 2 \text{CO}_2 \text{(g)} + 2 \text{H}_2\text{O (l)} \rightarrow \text{CH}_3\text{COOH (l)} + 2 \text{O}_2 \text{(g)} \]
The enthalpy change for this reaction will be:
\[ \Delta H = +210.0 \text{ kcal} \]

### Step 4: Adjust the second reaction
Since we need two moles of water, we will multiply the second reaction by 2:
\[ 2 \text{H}_2 \text{(g)} + \text{O}_2 \text{(g)} \rightarrow 2 \text{H}_2\text{O (l)} \]
The enthalpy change for this reaction will be:
\[ \Delta H = 2 \times (-68.0) = -136.0 \text{ kcal} \]

### Step 5: Adjust the first reaction
We will also multiply the first reaction by 2 to match the number of moles of CO₂ produced:
\[ 2 \text{C (s)} + 2 \text{O}_2 \text{(g)} \rightarrow 2 \text{CO}_2 \text{(g)} \]
The enthalpy change for this reaction will be:
\[ \Delta H = 2 \times (-94.0) = -188.0 \text{ kcal} \]

### Step 6: Combine all reactions
Now we can combine all the adjusted reactions:
1. \( 2 \text{C (s)} + 2 \text{O}_2 \text{(g)} \rightarrow 2 \text{CO}_2 \text{(g)}, \Delta H = -188.0 \text{ kcal} \)
2. \( 2 \text{H}_2 \text{(g)} + \text{O}_2 \text{(g)} \rightarrow 2 \text{H}_2\text{O (l)}, \Delta H = -136.0 \text{ kcal} \)
3. \( 2 \text{CO}_2 \text{(g)} + 2 \text{H}_2\text{O (l)} \rightarrow \text{CH}_3\text{COOH (l)} + 2 \text{O}_2 \text{(g)}, \Delta H = +210.0 \text{ kcal} \)

### Step 7: Calculate the total enthalpy change
Now we can sum the enthalpy changes:
\[ \Delta H_{\text{formation}} = (-188.0) + (-136.0) + 210.0 \]
\[ \Delta H_{\text{formation}} = -114.0 \text{ kcal} \]

### Final Answer
Thus, the heat of formation of acetic acid is:
\[ \Delta H_f = -114.0 \text{ kcal} \]

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