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Reaction 2BaO(2)(s)hArr2BaO(s) + O(2)(g)...

Reaction `2BaO_(2)(s)hArr2BaO(s) + O_(2)(g), Delta H = +ve`. At equilibrium condition, pressure of `O_(2)` is depended on:

A

increased mass of `BaO_2`

B

increased mass of `BaO`

C

increased temperature of equilibrium

D

increased mass of `BaO_2` and `BaO` both

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The correct Answer is:
To solve the question regarding the reaction \(2BaO_2(s) \rightleftharpoons 2BaO(s) + O_2(g)\) with a positive \(\Delta H\), we need to analyze how the pressure of \(O_2\) at equilibrium is affected by various factors. ### Step-by-Step Solution: 1. **Understanding the Reaction**: The reaction involves the decomposition of barium peroxide (\(BaO_2\)) into barium oxide (\(BaO\)) and oxygen gas (\(O_2\)). The reaction is represented as: \[ 2BaO_2(s) \rightleftharpoons 2BaO(s) + O_2(g) \] Here, \(BaO_2\) and \(BaO\) are solids, while \(O_2\) is a gas. 2. **Identify the Factors Affecting Equilibrium**: According to Le Chatelier's principle, the equilibrium position can be affected by changes in concentration, temperature, and pressure. However, the pressure of a gas at equilibrium primarily depends on the concentration of that gas and temperature. 3. **Analyzing Each Option**: - **Option 1: Increased mass of \(BaO_2\)**: Since \(BaO_2\) is a solid, increasing its mass does not affect the pressure of \(O_2\) because solids do not contribute to the pressure of gases in equilibrium. Thus, this option is incorrect. - **Option 2: Increased mass of \(BaO\)**: Similar to \(BaO_2\), \(BaO\) is also a solid. Therefore, increasing its mass will not affect the pressure of \(O_2\) at equilibrium. This option is also incorrect. - **Option 3: Increased temperature of equilibrium**: Since the reaction has a positive \(\Delta H\), it is endothermic. According to Le Chatelier's principle, increasing the temperature will shift the equilibrium to the right, producing more \(O_2\) and thus increasing its pressure. This option is correct. - **Option 4: Increased mass of \(BaO_2\) and \(BaO\)**: As previously discussed, neither solid affects the pressure of the gas. Therefore, this option is incorrect. 4. **Conclusion**: The only factor that affects the pressure of \(O_2\) at equilibrium is the increased temperature. Hence, the correct answer is: \[ \text{Correct Option: Increased temperature of equilibrium} \]
To solve the question regarding the reaction \(2BaO_2(s) \rightleftharpoons 2BaO(s) + O_2(g)\) with a positive \(\Delta H\), we need to analyze how the pressure of \(O_2\) at equilibrium is affected by various factors. ### Step-by-Step Solution: 1. **Understanding the Reaction**: The reaction involves the decomposition of barium peroxide (\(BaO_2\)) into barium oxide (\(BaO\)) and oxygen gas (\(O_2\)). The reaction is represented as: \[ 2BaO_2(s) \rightleftharpoons 2BaO(s) + O_2(g) ...
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