ANS:C - 2

In a chemical reaction, the rate of reaction typically depends on the concentration of the reactants involved. This dependence is often expressed using a rate law equation. For a generic reaction: 𝑎𝐴+𝑏𝐵→𝑐𝐶+𝑑𝐷aA+bB→cC+dD The rate law equation might look like this: Rate=𝑘[𝐴]𝑥[𝐵]𝑦Rate=k[A]x[B]y Where:

• RateRate is the rate of the reaction,
• 𝑘k is the rate constant,
• [𝐴][A] and [𝐵][B] are the concentrations of reactants A and B, respectively,
• 𝑥x and 𝑦y are the reaction orders with respect to reactants A and B, respectively.

Given your scenario, the rate is said to become nine times when the concentration of 'X' (let's denote it as [𝑋][X]) is tripled. Let's denote the initial rate of reaction as 𝑅0R0​ and the initial concentration of 'X' as [𝑋]0[X]0​. When the concentration of 'X' is tripled, it becomes 3[𝑋]03[X]0​. According to the information provided, the new rate, let's call it 𝑅′R′, is nine times the initial rate 𝑅0R0​. So, mathematically, we can express this relationship as: 𝑅′=9𝑅0R′=9R0​ If the rate law follows a power dependence on the concentration of 'X', then we can write: 𝑅′=𝑘([𝑋]0×3)𝑥R′=k([X]0​×3)x 9𝑅0=𝑘(3[𝑋]0)𝑥9R0​=k(3[X]0​)x 9=3𝑥9=3x Now, solving for 𝑥x, we find that 𝑥=2x=2. However, this doesn't match the given scenario where the rate becomes nine times. This means that the rate doesn't increase proportionally with the square of the concentration of 'X'. Let's try another scenario. If the rate becomes nine times when the concentration of 'X' is tripled, then it suggests the rate is directly proportional to the cube of the concentration of 'X'. So, mathematically: 𝑅′=𝑘([𝑋]0×3)𝑥R′=k([X]0​×3)x 9𝑅0=𝑘(3[𝑋]0)𝑥9R0​=k(3[X]0​)x 9=3𝑥9=3x 𝑥=2x=2 This suggests that the reaction order with respect to 'X' is 3. Hence, it's a third-order reaction with respect to 'X'.