Question

at 8. (R in L.atm.K⁻¹ mole⁻¹). (a) The given reaction will be exothermic in nature due to the formation of three X - Y bonds from the gaseous atoms. The reaction is also accompanied with the decrease in the gaseous species (i.e. $\Delta n$ is negative). Hence, the reaction will be affected by both temperature and pressure.

Answer 1

The statement (a) is correct.

Answer 2

Explanation:

The given statement (a) describes a reaction involving the formation of chemical bonds from gaseous atoms and discusses how temperature and pressure affect it. Let's analyze each part:

1. Exothermic nature due to bond formation: The formation of chemical bonds from isolated atoms (which are generally at a higher energy state) is an energy-releasing process. This means the enthalpy change ($\Delta H$) for bond formation is negative, making the reaction exothermic. This part of the statement is correct.

2. Decrease in gaseous species ($\Delta n < 0$): When multiple gaseous atoms combine to form fewer gaseous molecules, the total number of moles of gas decreases. For example, if 1 mole of X atoms and 3 moles of Y atoms combine to form 1 mole of XY₃ molecules (assuming XY₃ is gaseous), the change in the number of moles of gas ($\Delta n$) is (moles of gaseous products) - (moles of gaseous reactants) = $1 - (1+3) = -3$. A decrease in gaseous species is typical for such association reactions. This part of the statement is plausible and consistent with the reaction description.

3. Effect of temperature: The equilibrium constant ($K$) of a reaction is temperature-dependent. For an exothermic reaction ($\Delta H < 0$), increasing the temperature shifts the equilibrium towards the reactants (according to Le Chatelier's principle or the van't Hoff equation: $\frac{d(\ln K)}{dT} = \frac{\Delta H^\circ}{RT^2}$). Conversely, decreasing the temperature shifts it towards the products. Thus, the equilibrium is affected by temperature. This part of the statement is correct.

4. Effect of pressure: For a gaseous reaction where the number of moles of gas changes ($\Delta n \neq 0$), changing the total pressure affects the equilibrium position according to Le Chatelier's principle. If $\Delta n < 0$ (decrease in moles of gas), increasing the pressure shifts the equilibrium towards the side with fewer moles of gas, which is the product side in this case. Decreasing the pressure shifts it towards the reactant side. Thus, the equilibrium is affected by pressure. This part of the statement is correct.

Since all parts of the reasoning presented in the statement are consistent with chemical principles, the overall statement (a) is correct.