For `CaCO_(3)(s)rarr CaO(s)+CO_(2)(g)` at `977^(@)C, Delta H = 174` KJ/mol , then `Delta E` is :-

Options

A. 160 KJ
B. 163.6 KJ
C. 186.4 KJ
D. 180 KJ

Answer

To find the value of ΔE for the reaction:

\[ \text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g) \]

at a temperature of \( 977^\circ C \) with \( \Delta H = 174 \, \text{kJ/mol} \), we can follow these steps:

### Step 1: Identify the change in the number of moles of gas (ΔNG)
In the given reaction, we have:
- Reactants: 1 mole of CaCO₃ (solid)
- Products: 1 mole of CaO (solid) and 1 mole of CO₂ (gas)

The change in the number of moles of gas (ΔNG) is calculated as:
\[ \Delta NG = \text{moles of gaseous products} - \text{moles of gaseous reactants} \]
\[ \Delta NG = 1 - 0 = 1 \]

### Step 2: Use the relationship between ΔH and ΔE
The relationship between enthalpy change (ΔH) and internal energy change (ΔE) is given by:
\[ \Delta H = \Delta E + \Delta NG \cdot R \cdot T \]
Where:
- R = 8.314 J/(mol·K) = 0.008314 kJ/(mol·K)
- T = temperature in Kelvin

### Step 3: Convert the temperature to Kelvin
To convert the temperature from Celsius to Kelvin:
\[ T(K) = T(°C) + 273 \]
\[ T = 977 + 273 = 1250 \, K \]

### Step 4: Substitute the values into the equation
Now we can substitute the known values into the equation:
\[ 174 \, \text{kJ} = \Delta E + (1) \cdot (0.008314 \, \text{kJ/(mol·K)}) \cdot (1250 \, K) \]

### Step 5: Calculate the term involving R and T
Calculate \( R \cdot T \):
\[ R \cdot T = 0.008314 \, \text{kJ/(mol·K)} \cdot 1250 \, K = 10.3925 \, \text{kJ} \]

### Step 6: Solve for ΔE
Now we can solve for ΔE:
\[ 174 \, \text{kJ} = \Delta E + 10.3925 \, \text{kJ} \]
\[ \Delta E = 174 \, \text{kJ} - 10.3925 \, \text{kJ} \]
\[ \Delta E = 163.6075 \, \text{kJ} \]

### Step 7: Round the answer
Rounding to two decimal places, we get:
\[ \Delta E \approx 163.61 \, \text{kJ} \]

### Final Answer
Thus, the value of ΔE is approximately:
\[ \Delta E \approx 163.60 \, \text{kJ} \]

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