Question

The reaction CH₃–CH₂–NO₂+OH–→CH₃–CH–NO₂+H₂O obeys the rate law for pseudo first order kinetics in the presence of a large excess of hydroxide ion. If 1% of nitro ethane undergoes reaction in half a minute when the reactant concentration is 0.002 M, What is the pseudo first order rate constant ?

Answer 1

The pseudo first-order rate is constant for the reaction given is calculated using the formula k = ln(2) / t(1/2), resulting in a value of 0.0231 sec−¹.

Step-by-step explanation:

The reaction CH₃–CH₂–NO₂ + OH− → CH₃–CH–NO₂ + H₂O is described as following pseudo-first-order kinetics in a large excess of hydroxide ion. In this scenario, to find the pseudo-first-order rate constant, one can use the formula for first-order reactions: k = ln(2) / t(1/2), where t(1/2) is the half-life time.

Given that 1% of nitroethane reacts in 30 seconds with a concentration of 0.002 M, we first convert the percentage reacted to a fraction (0.01) and use the half-life formula to find the rate constant.

Using the formula k = ln(2) / t(1/2), we have:

k = ln(2) / 30 seconds
k = 0.693 / 30 seconds
k = 0.0231 sec−¹

The pseudo-first-order rate constant (k) for the reaction is therefore 0.0231 sec−¹.