Given `C_(("graphite"))+O_(2)(g)toCO_(2)(g),`
`Delta_(r)H^(0)=-393.5kJ" "mol^(-1)`
`H_(2)(g)=+(1)/(2)O_(2)(g)toH_(2)O(1),`
`Delta_(r)H^(0)=-285.8" kJ "mol^(-1)`
`CO_(2)(g)+2H_(2)O(1)toCH_(4)(g)+2O_(2)(g)`,
`Delta_(r)H^(0)=+890.3kJ" "mol^(-1)`
Based on the above thermochemical equations, the value of `Delta_(r)H^(0)` at at 298 K for the reaction
`C_(("graphite"))+2H_(2)(g)toCH_(4)(g)` will be:

Options

A. `+144.0" kJ "mol^(-1)`
B. `-74.8kJ " "mol^(-1)`
C. `-144.0" kJ "mol^(-1)`
D. `+74.8" kJ "mol^(-1)`

Answer

To find the value of ΔrH° at 298 K for the reaction:

\[ C_{\text{(graphite)}} + 2H_2(g) \rightarrow CH_4(g) \]

we will manipulate the provided thermochemical equations and their enthalpy changes.

### Step 1: Write down the given reactions and their ΔrH° values.

1. \( C_{\text{(graphite)}} + O_2(g) \rightarrow CO_2(g) \)
\( \Delta_rH^{0} = -393.5 \, \text{kJ/mol} \) (Equation 1)

2. \( H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l) \)
\( \Delta_rH^{0} = -285.8 \, \text{kJ/mol} \) (Equation 2)

3. \( CO_2(g) + 2H_2O(l) \rightarrow CH_4(g) + 2O_2(g) \)
\( \Delta_rH^{0} = +890.3 \, \text{kJ/mol} \) (Equation 3)

### Step 2: Manipulate the equations to derive the target reaction.

To obtain the target reaction \( C_{\text{(graphite)}} + 2H_2(g) \rightarrow CH_4(g) \), we will:

- Use Equation 1 as it is.
- Use Equation 2, but we need 2 moles of \( H_2 \), so we will multiply it by 2:

\[ 2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \]
\( \Delta_rH^{0} = 2 \times (-285.8) = -571.6 \, \text{kJ/mol} \) (Modified Equation 2)

- Use Equation 3 as it is.

### Step 3: Add the manipulated equations together.

Now we will add the equations:

1. \( C_{\text{(graphite)}} + O_2(g) \rightarrow CO_2(g) \)
2. \( 2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \)
3. \( CO_2(g) + 2H_2O(l) \rightarrow CH_4(g) + 2O_2(g) \)

Combining these gives:

\[ C_{\text{(graphite)}} + O_2(g) + 2H_2(g) + O_2(g) + CO_2(g) + 2H_2O(l) \rightarrow CO_2(g) + 2H_2O(l) + CH_4(g) + 2O_2(g) \]

### Step 4: Cancel out the common species.

On simplifying, we cancel out \( CO_2(g) \) and \( 2H_2O(l) \) from both sides and also \( O_2(g) \):

This gives us:

\[ C_{\text{(graphite)}} + 2H_2(g) \rightarrow CH_4(g) \]

### Step 5: Calculate the overall ΔrH°.

Now we can calculate the overall ΔrH°:

\[ \Delta_rH^{0} = \Delta_rH^{0}_{\text{(1)}} + \Delta_rH^{0}_{\text{(2)}} + \Delta_rH^{0}_{\text{(3)}} \]

\[ \Delta_rH^{0} = (-393.5) + (-571.6) + (890.3) \]

Calculating this:

\[ \Delta_rH^{0} = -393.5 - 571.6 + 890.3 = -74.8 \, \text{kJ/mol} \]

### Final Answer:

The value of \( \Delta_rH^{0} \) for the reaction \( C_{\text{(graphite)}} + 2H_2(g) \rightarrow CH_4(g) \) is:

\[ \Delta_rH^{0} = -74.8 \, \text{kJ/mol} \]

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