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. +144kj/mol
B. -74.8kj/mol
C. -144kj/mol
D. +74.8kj/mol

Answer

To find the value of \( \Delta_r H^{0} \) at 298 K for the reaction

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

we will use the provided thermochemical equations and apply Hess's law.

### Step 1: Write down the given reactions and their enthalpy changes.

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

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

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

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

To derive the target reaction \( C_{\text{(graphite)}} + 2H_2(g) \to CH_4(g) \), we will manipulate the given reactions as follows:

- **From Equation 1**, we keep it as is:
\[ C_{\text{(graphite)}} + O_2(g) \to CO_2(g) \quad \text{(1)} \]

- **From Equation 2**, we need to multiply it by 2 to match the 2 moles of \( H_2 \) in our target reaction:
\[ 2H_2(g) + O_2(g) \to 2H_2O(l) \quad \Delta_r H^{0} = 2 \times (-285.8) = -571.6 \, \text{kJ/mol} \quad \text{(2)} \]

- **From Equation 3**, we reverse the reaction to get \( CH_4(g) \) on the left side:
\[ CH_4(g) + 2O_2(g) \to CO_2(g) + 2H_2O(l) \quad \Delta_r H^{0} = -890.3 \, \text{kJ/mol} \quad \text{(3)} \]

### Step 3: Combine the manipulated equations.

Now we will add the manipulated equations:

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

Combining these, we have:

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

Now, cancel out the common species:

- \( O_2(g) \) and \( CO_2(g) \) cancel out.

This gives us:

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

### Step 4: Calculate \( \Delta_r H^{0} \) for the target reaction.

Now we can calculate \( \Delta_r H^{0} \) for the target reaction:

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

Calculating this gives:

\[ \Delta_r H^{0} = -393.5 - 571.6 - 890.3 = -1855.4 \, \text{kJ/mol} \]

### Step 5: Final calculation

Now we need to add the enthalpy changes:

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

Calculating this:

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

### Final Answer

Thus, the value of \( \Delta_r H^{0} \) at 298 K for the reaction \( C_{\text{(graphite)}} + 2H_2(g) \to CH_4(g) \) is:

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

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