Ethyl chloride `(C_(2)H_(5)Cl)` , is prepared by reaction of ethylene with hydrogen chloride`:`
`C_(2)H_(4)(g)+HCl(g) rarr C_(2)H_(5)Cl(g)`
`DeltaH=-72.3kJ//mol`
What is the value of `DeltaE (` in `kJ)`, if `98g` of ethylene and `109.5g` of `HCl` are allowed to react at `300K`

Options

A. `-64.81`
B. `-190.71`
C. `-209.41`
D. `-224.38`

Explanation

To find the value of ΔE for the reaction of ethylene with hydrogen chloride, we can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced equation for the reaction is:
\[ C_2H_4(g) + HCl(g) \rightarrow C_2H_5Cl(g) \]

### Step 2: Calculate the number of moles of reactants
We need to calculate the number of moles of ethylene (C₂H₄) and hydrogen chloride (HCl) using their respective weights and molecular weights.

- **Molecular weight of ethylene (C₂H₄)**:
\[ C: 12.01 \, \text{g/mol} \times 2 + H: 1.008 \, \text{g/mol} \times 4 = 28.06 \, \text{g/mol} \]

- **Number of moles of ethylene**:
\[ \text{Moles of } C_2H_4 = \frac{\text{mass}}{\text{molecular weight}} = \frac{98 \, \text{g}}{28.06 \, \text{g/mol}} \approx 3.49 \, \text{moles} \]

- **Molecular weight of hydrogen chloride (HCl)**:
\[ H: 1.008 \, \text{g/mol} + Cl: 35.45 \, \text{g/mol} = 36.46 \, \text{g/mol} \]

- **Number of moles of HCl**:
\[ \text{Moles of HCl} = \frac{109.5 \, \text{g}}{36.46 \, \text{g/mol}} \approx 3.01 \, \text{moles} \]

### Step 3: Identify the limiting reactant
From the balanced equation, we see that 1 mole of C₂H₄ reacts with 1 mole of HCl. Since we have approximately 3.49 moles of C₂H₄ and 3.01 moles of HCl, HCl is the limiting reactant.

### Step 4: Calculate the change in moles (ΔN)
Using the stoichiometry of the reaction:
- Moles of products = 3 moles (from 3 moles of HCl)
- Moles of reactants = 3 moles (C₂H₄) + 3 moles (HCl) = 6 moles

Thus,
\[ \Delta N = \text{Moles of products} - \text{Moles of reactants} = 3 - 6 = -3 \]

### Step 5: Use the formula to calculate ΔE
The relationship between ΔH and ΔE is given by:
\[ \Delta H = \Delta E - \Delta N_{gr}RT \]
Rearranging gives:
\[ \Delta E = \Delta H + \Delta N_{gr}RT \]

Where:
- ΔH = -72.3 kJ/mol
- R = 8.314 J/(mol·K) = 0.008314 kJ/(mol·K)
- T = 300 K

Now substituting the values:
\[ \Delta E = -72.3 + (-3)(0.008314)(300) \]

Calculating ΔNgrRT:
\[ \Delta N_{gr}RT = -3 \times 0.008314 \, \text{kJ/(mol·K)} \times 300 \, \text{K} \]
\[ = -7.48 \, \text{kJ} \]

Now substituting this back into the equation for ΔE:
\[ \Delta E = -72.3 - 7.48 \]
\[ \Delta E = -79.78 \, \text{kJ} \]

### Final Answer
Thus, the value of ΔE is approximately:
\[ \Delta E \approx -79.78 \, \text{kJ} \]

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