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. gt 80.25kJ
C. lt 80.25 kJ
D. 160.50 kJ

Explanation

To solve the problem of determining the heat change at constant pressure for the decomposition of silver oxide, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Reaction**: The decomposition of silver oxide (Ag2O) can be represented as:
\[ 2 \text{Ag}_2\text{O} \rightarrow 4 \text{Ag} + \text{O}_2 \]
This reaction shows that silver oxide decomposes into silver and oxygen gas.

2. **Identify Given Information**: We are given that the heat change at constant volume (\( \Delta U \)) for this reaction is 80.25 kJ.

3. **Relate Internal Energy and Enthalpy**: The relationship between the change in enthalpy (\( \Delta H \)) and the change in internal energy (\( \Delta U \)) at constant pressure is given by the equation:
\[ \Delta H = \Delta U + P \Delta V \]
where \( P \) is the pressure and \( \Delta V \) is the change in volume.

4. **Calculate Change in Moles of Gas**: To find \( P \Delta V \), we need to determine the change in the number of moles of gas (\( \Delta n_g \)):
- **Products**: In the products, we have 1 mole of \( O_2 \) (gas).
- **Reactants**: In the reactants, we have 0 moles of gas (since \( Ag_2O \) is a solid).
- Therefore, \( \Delta n_g = n_{\text{products}} - n_{\text{reactants}} = 1 - 0 = 1 \).

5. **Use Ideal Gas Law**: According to the ideal gas law, we can express \( P \Delta V \) in terms of \( \Delta n_g \):
\[ P \Delta V = \Delta n_g R T \]
where \( R \) is the gas constant and \( T \) is the temperature in Kelvin.

6. **Substitute Values**: Now we can substitute \( \Delta n_g \) into the enthalpy equation:
\[ \Delta H = \Delta U + \Delta n_g R T \]
Substituting the known values:
\[ \Delta H = 80.25 \text{ kJ} + (1) R T \]

7. **Conclusion**: Since \( R \) and \( T \) are both positive constants, \( \Delta H \) will be greater than \( 80.25 \text{ kJ} \). Thus, we conclude:
\[ \Delta H > 80.25 \text{ kJ} \]

### Final Answer:
The heat change at constant pressure for the decomposition of silver oxide will be greater than 80.25 kJ.