Determine `Delta U^(@) ` at 300K for the following reaction using the listed enthalpies of reaction :
`4CO(g)+8H_(2)(g)to3CH_(4)(g)+CO_(2)(g)+2H_(2)O(l)`
`C("graphite")+(1)/(2)O_(2)(g)to CO(g), DeltaH_(1)^(@)=-110.5KJ`
` CO(g)(1)/(2)O_(2)(g)toCO_(2)(g), DeltaH_(2)^(@)=-282.9KJ`
`H_(2)(g)+(1)/(2)O_(2)(g)to H_(2)O(l), DeltaH_(3)^(@)=-285.8KJ`
`C("graphite")+2H_(2)(g)to CH_(4)(g), DeltaH_(4)^(@)=-74.8KJ`

Options

A. `-653.5` kJ
B. `-686.2` kKJ
C. `-747.4` kJ
D. None of these

Explanation

To determine the change in internal energy (ΔU) at 300 K for the reaction:

\[ 4CO(g) + 8H_2(g) \rightarrow 3CH_4(g) + CO_2(g) + 2H_2O(l) \]

we will use the provided enthalpies of reaction and apply the relationship between ΔH and ΔU.

### Step-by-Step Solution:

1. **Identify the Reactions and Their Enthalpies**:
We have the following reactions with their respective enthalpies:
- Reaction 1:
\[ C(s) + \frac{1}{2}O_2(g) \rightarrow CO(g), \quad \Delta H_1^\circ = -110.5 \, \text{kJ} \]
- Reaction 2:
\[ CO(g) + \frac{1}{2}O_2(g) \rightarrow CO_2(g), \quad \Delta H_2^\circ = -282.9 \, \text{kJ} \]
- Reaction 3:
\[ H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l), \quad \Delta H_3^\circ = -285.8 \, \text{kJ} \]
- Reaction 4:
\[ C(s) + 2H_2(g) \rightarrow CH_4(g), \quad \Delta H_4^\circ = -74.8 \, \text{kJ} \]

2. **Reverse and Adjust Reactions**:
To obtain the target reaction, we need to manipulate the given reactions:
- Reverse Reaction 1 and multiply by 3:
\[ 3CO(g) \rightarrow 3C(s) + \frac{3}{2}O_2(g), \quad \Delta H = +3 \times 110.5 = +331.5 \, \text{kJ} \]
- Use Reaction 2 as it is:
\[ CO(g) + \frac{1}{2}O_2(g) \rightarrow CO_2(g), \quad \Delta H = -282.9 \, \text{kJ} \]
- Multiply Reaction 3 by 2:
\[ 2H_2(g) + O_2(g) \rightarrow 2H_2O(l), \quad \Delta H = 2 \times -285.8 = -571.6 \, \text{kJ} \]
- Multiply Reaction 4 by 3:
\[ 3C(s) + 6H_2(g) \rightarrow 3CH_4(g), \quad \Delta H = 3 \times -74.8 = -224.4 \, \text{kJ} \]

3. **Combine the Reactions**:
Now, we add the modified reactions:
\[ 3CO(g) + CO(g) + 2H_2(g) + 3C(s) + 6H_2(g) \rightarrow 3C(s) + 3CH_4(g) + CO_2(g) + 2H_2O(l) \]
This simplifies to:
\[ 4CO(g) + 8H_2(g) \rightarrow 3CH_4(g) + CO_2(g) + 2H_2O(l) \]

4. **Calculate Total ΔH**:
Now, we can find ΔH for the overall reaction:
\[ \Delta H = 331.5 - 282.9 - 571.6 - 224.4 = -747.4 \, \text{kJ} \]

5. **Calculate ΔU**:
Use the formula:
\[ \Delta U = \Delta H - \Delta n_g RT \]
Where:
- \( \Delta n_g = \text{moles of gaseous products} - \text{moles of gaseous reactants} \)
- Products: \( 3CH_4(g) + CO_2(g) \) gives 4 moles of gas.
- Reactants: \( 4CO(g) + 8H_2(g) \) gives 12 moles of gas.
- Thus, \( \Delta n_g = 4 - 12 = -8 \).

Now, substituting values:
\[ R = 8.314 \, \text{J/(mol K)} = 0.008314 \, \text{kJ/(mol K)} \]
\[ T = 300 \, \text{K} \]
\[ \Delta U = -747.4 - (-8)(0.008314)(300) \]
\[ \Delta U = -747.4 + 19.98 = -727.42 \, \text{kJ} \]

### Final Answer:
\[ \Delta U \approx -727.4 \, \text{kJ} \]