Ethyl chloride `(C_(2)H_(5)Cl)`, is prepared by reaction of ethylene with hydrogen chloride :
`C_(2)H_(4)(g)+HCl(g)toC_(2)H_(5)cl(g)" "DeltaH=-72.3kJ`
What is the value of `DeltaE` (in kJ), if 70 g of ethylene and 73 g of HCL are allowed to react at 300 K.

Options

A. `-69.8`
B. `-180.75`
C. `-174.5`
D. `-139.6`

Answer

To find the value of `ΔE` for the reaction of ethylene `(C2H4)` with hydrogen chloride `(HCl)` to produce ethyl chloride `(C2H5Cl)`, we can follow these steps:

### Step 1: Calculate the number of moles of ethylene and hydrogen chloride.
1. **Given mass of ethylene (C2H4)** = 70 g
- Molar mass of C2H4 = 2(12) + 4(1) = 28 g/mol
- Number of moles of C2H4 = mass / molar mass = 70 g / 28 g/mol = 2.5 moles

2. **Given mass of hydrogen chloride (HCl)** = 73 g
- Molar mass of HCl = 1 + 35.5 = 36.5 g/mol
- Number of moles of HCl = mass / molar mass = 73 g / 36.5 g/mol = 2 moles

### Step 2: Identify the limiting reagent.
- The balanced reaction is:
\[ C2H4(g) + HCl(g) \rightarrow C2H5Cl(g) \]
- From the stoichiometry of the reaction, 1 mole of C2H4 reacts with 1 mole of HCl.
- We have 2.5 moles of C2H4 and 2 moles of HCl. Since HCl is present in a lesser amount, it is the limiting reagent.

### Step 3: Calculate the change in the number of moles of gas (ΔNg).
- **Reactants**: 1 mole of C2H4 + 1 mole of HCl = 2 moles of reactants
- **Products**: 1 mole of C2H5Cl = 1 mole of products
- ΔNg = moles of products - moles of reactants = 1 - 2 = -1

### Step 4: Use the relationship between ΔH and ΔE.
- The relationship is given by:
\[ ΔH = ΔE + ΔNg \cdot R \cdot T \]
- Given ΔH = -72.3 kJ, R = 8.314 J/(mol·K) = 0.008314 kJ/(mol·K), and T = 300 K.

### Step 5: Substitute values into the equation.
- Substitute ΔH, ΔNg, R, and T into the equation:
\[ -72.3 = ΔE + (-1) \cdot (0.008314) \cdot (300) \]
- Calculate the term involving R and T:
\[ -72.3 = ΔE - 2.4942 \]
- Rearranging gives:
\[ ΔE = -72.3 + 2.4942 = -69.8058 \text{ kJ} \]

### Step 6: Finalize the answer.
- Since we are looking for ΔE for the reaction involving 2 moles of HCl (as it is the limiting reagent), we multiply by 2:
\[ ΔE \text{ for 2 moles} = -69.8058 \times 2 = -139.6116 \text{ kJ} \]
- Rounding gives:
\[ ΔE \approx -139.6 \text{ kJ} \]

### Final Answer:
\[ \Delta E \approx -139.6 \text{ kJ} \]

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