`DeltaH" for "CaCO_(3(s))toCaO_((s))+CO_(2(g))" is "176kJmol^(-1)` at 1240 K. The `DeltaU` for the change is equal to :

Options

A. 160 kJ
B. 165.6 kJ
C. 186.3 kJ
D. 180.0 kJ

Answer

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

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

given that ΔH = 176 kJ/mol at 1240 K, we can use the relationship between ΔH and ΔU:

\[ \Delta H = \Delta U + \Delta N_g RT \]

where:
- ΔN_g = change in the number of moles of gas (products - reactants)
- R = universal gas constant (8.314 J/(mol·K))
- T = temperature in Kelvin

### Step 1: Calculate ΔN_g
In the given reaction:
- On the left side (reactants), we have 1 mole of solid CaCO₃ (which does not count towards ΔN_g).
- On the right side (products), we have 1 mole of gas (CO₂).

Thus, the change in the number of moles of gas (ΔN_g) is:
\[ \Delta N_g = 1 - 0 = 1 \]

### Step 2: Convert R to kJ
Since we want our final answer in kJ, we need to convert R from J to kJ:
\[ R = 8.314 \, \text{J/(mol·K)} = \frac{8.314}{1000} \, \text{kJ/(mol·K)} = 0.008314 \, \text{kJ/(mol·K)} \]

### Step 3: Substitute values into the equation
Now we can substitute ΔH, ΔN_g, R, and T into the equation:
\[ \Delta H = \Delta U + \Delta N_g RT \]
Substituting the known values:
\[ 176 \, \text{kJ/mol} = \Delta U + (1)(0.008314 \, \text{kJ/(mol·K)})(1240 \, \text{K}) \]

### Step 4: Calculate the term ΔN_g RT
Calculating the right-hand side:
\[ \Delta N_g RT = 1 \times 0.008314 \times 1240 = 10.3093 \, \text{kJ/mol} \]

### Step 5: Rearranging to find ΔU
Now we can rearrange the equation to solve for ΔU:
\[ \Delta U = \Delta H - \Delta N_g RT \]
Substituting the values:
\[ \Delta U = 176 \, \text{kJ/mol} - 10.3093 \, \text{kJ/mol} = 165.6907 \, \text{kJ/mol} \]

### Step 6: Rounding the answer
Rounding to three significant figures, we get:
\[ \Delta U \approx 165.7 \, \text{kJ/mol} \]

### Final Answer
Thus, the change in internal energy (ΔU) for the reaction is approximately:
\[ \Delta U \approx 165.6 \, \text{kJ/mol} \]