The internal energy change `( Delta U ) ` for the reaction `CH_(4)(g) + 2O_(2)(g) rarr CO_(2)(g) +2H_(2)O(l) ` is - 885kJ `mol^(-1)` at 298 K . What is `DeltaH` at298 K ?

Answer

To find the change in enthalpy (ΔH) for the given reaction, we can use the relationship between internal energy change (ΔU) and enthalpy change (ΔH). The relevant formula is:

\[ \Delta H = \Delta U + RT \Delta N_g \]

Where:
- \( \Delta H \) = change in enthalpy
- \( \Delta U \) = change in internal energy (given as -885 kJ/mol)
- \( R \) = universal gas constant (8.314 J/mol·K)
- \( T \) = temperature in Kelvin (given as 298 K)
- \( \Delta N_g \) = change in the number of moles of gas (moles of gaseous products - moles of gaseous reactants)

### Step 1: Identify the given values
- \( \Delta U = -885 \, \text{kJ/mol} \)
- \( T = 298 \, \text{K} \)
- \( R = 8.314 \, \text{J/mol·K} = 0.008314 \, \text{kJ/mol·K} \) (conversion from J to kJ)

### Step 2: Calculate \( \Delta N_g \)
The reaction is:
\[ CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l) \]

- Gaseous products: 1 mole of \( CO_2 \)
- Gaseous reactants: 1 mole of \( CH_4 \) + 2 moles of \( O_2 \) = 3 moles

Thus,
\[ \Delta N_g = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} = 1 - 3 = -2 \]

### Step 3: Substitute values into the ΔH formula
Now we can substitute the values into the formula:
\[ \Delta H = \Delta U + RT \Delta N_g \]
\[ \Delta H = -885 \, \text{kJ/mol} + (0.008314 \, \text{kJ/mol·K}) \times (298 \, \text{K}) \times (-2) \]

### Step 4: Calculate \( RT \Delta N_g \)
Calculating \( RT \Delta N_g \):
\[ RT \Delta N_g = 0.008314 \times 298 \times (-2) = -4.95 \, \text{kJ/mol} \]

### Step 5: Final calculation of ΔH
Now substituting back:
\[ \Delta H = -885 \, \text{kJ/mol} - 4.95 \, \text{kJ/mol} = -889.95 \, \text{kJ/mol} \]

### Final Answer
\[ \Delta H \approx -890 \, \text{kJ/mol} \]