If the heat change at constant volume for decomposition of silver oxide is 80.25 kJ, what will be the heat change at constant pressure ?

Options

A. 80.25 kJ
B. >80.25kJ
C. <80.25 kJ
D. 160.50 kJ

Explanation

To solve the problem, we need to find the heat change at constant pressure (ΔH) given the heat change at constant volume (ΔU) for the decomposition of silver oxide (Ag2O).

### Step-by-Step Solution:

1. **Write the Decomposition Reaction**:
The decomposition of silver oxide can be represented by the following chemical equation:
\[ 2 \text{Ag}_2\text{O (s)} \rightarrow 4 \text{Ag (s)} + \text{O}_2 (g) \]

2. **Identify Given Values**:
- Heat change at constant volume (ΔU) = 80.25 kJ

3. **Understand the Relationship Between ΔH and ΔU**:
The relationship between the heat change at constant pressure (ΔH) and the heat change at constant volume (ΔU) is given by the equation:
\[ \Delta H = \Delta U + \Delta N_g RT \]
where:
- ΔN_g = Change in the number of moles of gas
- R = Universal gas constant (8.314 J/(mol·K))
- T = Temperature in Kelvin

4. **Calculate ΔN_g**:
ΔN_g is calculated as:
\[ \Delta N_g = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} \]
In our reaction, we have:
- Gaseous products: 0.5 moles of O2
- Gaseous reactants: 0 moles (since Ag2O is a solid)

Therefore,
\[ \Delta N_g = 0.5 - 0 = 0.5 \]

5. **Substitute Values into the Equation**:
Now we can substitute ΔU and ΔN_g into the equation:
\[ \Delta H = 80.25 \text{ kJ} + (0.5) \cdot (8.314 \text{ J/(mol·K)}) \cdot T \]
Note: Since ΔH is in kJ, we need to convert R to kJ:
\[ R = 0.008314 \text{ kJ/(mol·K)} \]
Thus, the equation becomes:
\[ \Delta H = 80.25 \text{ kJ} + (0.5) \cdot (0.008314 \text{ kJ/(mol·K)}) \cdot T \]

6. **Conclusion**:
The value of ΔH will depend on the temperature (T). However, we can conclude that ΔH will be greater than ΔU (which is 80.25 kJ) because of the additional term (0.5 RT). Therefore, ΔH > 80.25 kJ.

### Final Answer:
The heat change at constant pressure (ΔH) is greater than 80.25 kJ.