The value of `DeltaE` for combustion of 16 g of `CH_4` is -885389 at 298 K. The `DeltaH` combustion for `CH_4` in `Jmol^(-1)` at this temperature will be : (Given that, `R = 8.314 JK^(-1) mol^(-1)`)

Options

A. (a) `-55337`
B. (b) `-880430`
C. (c) `-885389`
D. (d) `-890344`

Explanation

To find the value of ΔH (enthalpy change) for the combustion of methane (CH₄) given ΔE (internal energy change), we can use the relationship between ΔH and ΔE, which is:

\[ \Delta H = \Delta E + \Delta N_g \cdot R \cdot T \]

Where:
- ΔE = -885389 J (given)
- R = 8.314 J/(K·mol) (given)
- T = 298 K (given)
- ΔN_g = change in the number of moles of gas

### Step 1: Calculate the number of moles of CH₄
Given that the molecular weight of CH₄ is 16 g/mol, we can calculate the number of moles of CH₄ in 16 g.

\[ \text{Moles of CH₄} = \frac{\text{mass}}{\text{molecular weight}} = \frac{16 \text{ g}}{16 \text{ g/mol}} = 1 \text{ mol} \]

### Step 2: Determine the change in the number of moles of gas (ΔN_g)
The balanced equation for the combustion of methane is:

\[ \text{CH₄ (g)} + 2 \text{O₂ (g)} \rightarrow \text{CO₂ (g)} + 2 \text{H₂O (l)} \]

In this reaction:
- Reactants: 1 mol of CH₄ + 2 mol of O₂ = 3 mol of gas
- Products: 1 mol of CO₂ = 1 mol of gas

Thus, the change in the number of moles of gas (ΔN_g) is:

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

### Step 3: Substitute values into the ΔH equation
Now we can substitute the values into the ΔH equation:

\[ \Delta H = \Delta E + \Delta N_g \cdot R \cdot T \]

Substituting the known values:

\[ \Delta H = -885389 \text{ J} + (-2) \cdot (8.314 \text{ J/(K·mol)}) \cdot (298 \text{ K}) \]

Calculating the second term:

\[ \Delta H = -885389 \text{ J} - 2 \cdot 8.314 \cdot 298 \]

Calculating \(2 \cdot 8.314 \cdot 298\):

\[ 2 \cdot 8.314 \cdot 298 = 4965.592 \text{ J} \]

Now substituting back:

\[ \Delta H = -885389 \text{ J} - 4965.592 \text{ J} = -890354.592 \text{ J} \]

### Step 4: Round the final answer
Rounding the final answer gives:

\[ \Delta H \approx -890344 \text{ J/mol} \]

### Final Answer
The enthalpy change for the combustion of CH₄ at 298 K is approximately:

\[ \Delta H \approx -890344 \text{ J/mol} \]