`X(g) + 2Y(g) rarr 2Z(g) + 3A(g)` The change in enthalpy at `27^(@)C` is 79.5 kJ. The value of `Delta`U is

Options

A. 74.5 kJ
B. 4.99 kJ
C. 79.5 kJ
D. 75.9 kJ

Answer

To find the change in internal energy (ΔU) for the reaction given by:

\[ X(g) + 2Y(g) \rightarrow 2Z(g) + 3A(g) \]

with a change in enthalpy (ΔH) of 79.5 kJ at 27°C, we can use the relationship between ΔH and ΔU:

\[ \Delta H = \Delta U + \Delta n_g R T \]

### Step 1: Convert temperature from Celsius to Kelvin
The temperature in Kelvin (T) can be calculated as:

\[ T(K) = T(°C) + 273 = 27 + 273 = 300 \, K \]

### Step 2: Calculate Δn_g
Δn_g is the change in the number of moles of gas, calculated as:

\[ \Delta n_g = \text{moles of gaseous products} - \text{moles of gaseous reactants} \]

From the reaction:
- Moles of gaseous products (2Z + 3A) = 2 + 3 = 5 moles
- Moles of gaseous reactants (X + 2Y) = 1 + 2 = 3 moles

Thus,

\[ \Delta n_g = 5 - 3 = 2 \]

### Step 3: Use the ideal gas constant (R)
The value of the universal gas constant (R) is:

\[ R = 8.314 \, \text{J/mol·K} = 0.008314 \, \text{kJ/mol·K} \]

### Step 4: Substitute values into the equation
Now we can substitute ΔH, Δn_g, R, and T into the equation:

\[ \Delta H = \Delta U + \Delta n_g R T \]

Substituting the known values:

\[ 79.5 \, \text{kJ} = \Delta U + (2)(0.008314 \, \text{kJ/mol·K})(300 \, K) \]

### Step 5: Calculate the term involving Δn_g, R, and T
Calculating the right-hand side:

\[ (2)(0.008314)(300) = 4.998 \, \text{kJ} \]

### Step 6: Rearrange to find ΔU
Now, substituting back into the equation:

\[ 79.5 \, \text{kJ} = \Delta U + 4.998 \, \text{kJ} \]

Rearranging gives:

\[ \Delta U = 79.5 \, \text{kJ} - 4.998 \, \text{kJ} \]

Calculating:

\[ \Delta U = 74.502 \, \text{kJ} \approx 74.5 \, \text{kJ} \]

### Final Answer
Thus, the change in internal energy (ΔU) is approximately:

\[ \Delta U \approx 74.5 \, \text{kJ} \]

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