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 solve the problem, we need to calculate the change in internal energy (ΔE) for the reaction of ethylene (C₂H₄) with hydrogen chloride (HCl) given the enthalpy change (ΔH) and the amounts of reactants. Here’s the step-by-step solution:

### 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: Identify the change in moles of gas (ΔNG)
To find ΔNG, we calculate the difference between the moles of gaseous products and the moles of gaseous reactants:
- Moles of products = 1 (C₂H₅Cl)
- Moles of reactants = 2 (C₂H₄ + HCl)

Thus,
\[ \Delta N_G = \text{Moles of products} - \text{Moles of reactants} = 1 - 2 = -1 \]

### Step 3: Use the relationship between ΔH and ΔE
The relationship between the change in enthalpy (ΔH) and the change in internal energy (ΔE) is given by:
\[ \Delta E = \Delta H - \Delta N_G RT \]
Where:
- \( R = 8.314 \, \text{J/mol·K} \) (universal gas constant)
- \( T = 300 \, \text{K} \)

### Step 4: Convert ΔH to the same units as ΔE
Given:
\[ \Delta H = -72.3 \, \text{kJ/mol} = -72300 \, \text{J/mol} \]

### Step 5: Calculate the term ΔNGRT
Now we calculate \( \Delta N_G RT \):
\[ \Delta N_G RT = (-1) \times 8.314 \, \text{J/mol·K} \times 300 \, \text{K} \]
\[ = -2494.2 \, \text{J/mol} \]

### Step 6: Substitute values into the ΔE equation
Now substitute the values into the equation for ΔE:
\[ \Delta E = -72300 \, \text{J/mol} - (-2494.2 \, \text{J/mol}) \]
\[ = -72300 + 2494.2 \]
\[ = -69805.8 \, \text{J/mol} \]

### Step 7: Convert ΔE back to kJ
Convert ΔE from J to kJ:
\[ \Delta E = -69.8058 \, \text{kJ/mol} \]

### Step 8: Calculate the moles of reactants
Now we need to determine the limiting reagent by calculating the moles of each reactant:
- Molar mass of C₂H₄ = 28 g/mol
\[ \text{Moles of C₂H₄} = \frac{98 \, \text{g}}{28 \, \text{g/mol}} = 3.5 \, \text{mol} \]

- Molar mass of HCl = 36.5 g/mol
\[ \text{Moles of HCl} = \frac{109.5 \, \text{g}}{36.5 \, \text{g/mol}} \approx 3 \, \text{mol} \]

### Step 9: Identify the limiting reagent
Since HCl is the limiting reagent (3 moles), we will use this to calculate the total ΔE for the reaction.

### Step 10: Calculate total ΔE for the reaction
The total ΔE for the reaction when 3 moles of HCl react is:
\[ \text{Total } \Delta E = 3 \times (-69.8058 \, \text{kJ/mol}) \]
\[ = -209.4174 \, \text{kJ} \]

### Final Answer
Thus, the value of ΔE for the reaction is approximately:
\[ \Delta E \approx -209.42 \, \text{kJ} \]

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