For the reaction
`C_(2)H_(4)(g)+3O_(2)(g) rarr 2CO_(2) (g) +2H_(2)O(l)`, Delta E=-1415 kJ`. The `DeltaH` at `27^(@)C` is

Options

A. `-1410 kJ`
B. `-1420 kJ`
C. `+1420 kJ`
D. `+1410 kJ`

Explanation

To find the value of ΔH at 27°C for the given reaction:

**Step 1: Write down the reaction and the given data.**
The reaction is:
\[ C_{2}H_{4}(g) + 3O_{2}(g) \rightarrow 2CO_{2}(g) + 2H_{2}O(l) \]
Given:
- ΔE = -1415 kJ
- Temperature (T) = 27°C = 300 K (since T in Kelvin = °C + 273)

**Step 2: Identify the relationship between ΔH and ΔE.**
The relationship is given by the equation:
\[ ΔH = ΔE + ΔN_{g}RT \]
Where:
- ΔH = change in enthalpy
- ΔE = change in internal energy
- ΔN_{g} = change in the number of moles of gas
- R = universal gas constant = 8.314 J/(mol·K) = 0.008314 kJ/(mol·K)
- T = temperature in Kelvin

**Step 3: Calculate ΔN_{g}.**
ΔN_{g} is calculated as:
\[ ΔN_{g} = \text{moles of gaseous products} - \text{moles of gaseous reactants} \]
From the reaction:
- Moles of gaseous products (2CO₂) = 2
- Moles of gaseous reactants (C₂H₄ + 3O₂) = 1 + 3 = 4

Thus,
\[ ΔN_{g} = 2 - 4 = -2 \]

**Step 4: Substitute the values into the equation for ΔH.**
Now we can substitute ΔE, ΔN_{g}, R, and T into the equation:
\[ ΔH = ΔE + ΔN_{g}RT \]
\[ ΔH = -1415 \, \text{kJ} + (-2) \times (0.008314 \, \text{kJ/(mol·K)}) \times (300 \, \text{K}) \]

**Step 5: Calculate the second term.**
Calculating the second term:
\[ -2 \times 0.008314 \times 300 = -4.989 \, \text{kJ} \]

**Step 6: Final calculation of ΔH.**
Now substituting this back into the equation:
\[ ΔH = -1415 \, \text{kJ} - 4.989 \, \text{kJ} \]
\[ ΔH = -1419.989 \, \text{kJ} \]

Rounding this to three significant figures gives:
\[ ΔH \approx -1420 \, \text{kJ} \]

**Final Answer:**
\[ ΔH = -1420 \, \text{kJ} \]

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