The following reaction are given :
`CH_(4(g))+2O_(2(g))toCO_(2(g))+2H_(2)O(g),DeltaH=-x`
`C_((s))+O_(2(g))toCO_(2(g)),2H_(2(g)),DeltaH=-y`
`H_(2(g))+(1)/(2)O_(2(g))toH_(2)O_((g)),2H_(2(g)),DeltaH=-z`
calculate the heat of formation of `CH_(4)`?

Options

A. x+y+z
B. y+2z-x
C. x-y-2z
D. none of the above

Explanation

To calculate the heat of formation of methane (CH₄) from the given reactions, we will follow these steps:

### Step 1: Write the formation reaction of CH₄
The standard formation reaction for methane (CH₄) from its elements in their standard states is:
\[ C_{(s)} + 2H_{2(g)} \rightarrow CH_{4(g)} \]
This reaction has an enthalpy change (ΔH) that we want to find.

### Step 2: Analyze the given reactions
We have three reactions with their respective enthalpy changes:
1. \( CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_{2}O_{(g)} \), ΔH = -x
2. \( C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)} \), ΔH = -y
3. \( H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow H_{2}O_{(g)} \), ΔH = -z

### Step 3: Reverse the first reaction
To get CH₄ on the right side, we reverse the first reaction:
\[ CO_{2(g)} + 2H_{2}O_{(g)} \rightarrow CH_{4(g)} + 2O_{2(g)} \]
Now, the enthalpy change becomes:
\[ \Delta H = +x \]

### Step 4: Keep the second reaction as is
The second reaction remains unchanged:
\[ C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)} \]
with ΔH = -y.

### Step 5: Modify the third reaction
We need to have 2H₂ on the left side. We can achieve this by multiplying the entire third reaction by 2:
\[ 2H_{2(g)} + O_{2(g)} \rightarrow 2H_{2}O_{(g)} \]
Now, the enthalpy change becomes:
\[ \Delta H = -2z \]

### Step 6: Combine the modified reactions
Now we will add the modified reactions together:
1. \( CO_{2(g)} + 2H_{2}O_{(g)} \rightarrow CH_{4(g)} + 2O_{2(g)} \) (ΔH = +x)
2. \( C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)} \) (ΔH = -y)
3. \( 2H_{2(g)} + O_{2(g)} \rightarrow 2H_{2}O_{(g)} \) (ΔH = -2z)

### Step 7: Cancel out similar terms
When we add these reactions, we can cancel out the terms:
- CO₂ from the first and second reactions
- 2H₂O from the first and third reactions
- O₂ from the first and third reactions

After cancellation, we are left with:
\[ C_{(s)} + 2H_{2(g)} \rightarrow CH_{4(g)} \]

### Step 8: Calculate the overall enthalpy change
The overall enthalpy change for the formation of CH₄ is:
\[ \Delta H = x - y - 2z \]

### Final Answer
Thus, the heat of formation of CH₄ is:
\[ \Delta H_{f} (CH₄) = x - y - 2z \]

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