Question

At a certain fixed temperature, the reaction below: A(g) 2B(g) AB2(g) is found to be first order in the concentration of A and zeroth order in the concentration of B. The reaction rate constant is 0.05s-1. If 2.00 moles of A and 4.00 moles of B are placed in a 1.00 liter container, how many seconds will elapse before the concentration of A has fallen to 0.30 moles/liter

Answer 1

By using the integrated rate law for a first-order reaction, we find that it takes approximately 37.942 seconds for the concentration of A to fall from 2.00 moles/liter to 0.30 moles/liter in the given reaction.

Step-by-step explanation:

To calculate the time it will take for the concentration of A to fall to 0.30 moles/liter in the given reaction where A(g) + 2B(g) → AB2(g), and the reaction is first order in A and zeroth order in B with a rate constant of 0.05s-1, we will use the integrated first-order rate law:

ln[A]t - ln[A]0 = -kt

Where:

[A]t is the final concentration of A (0.30 M),

[A]0 is the initial concentration of A (2.00 M),

k is the rate constant (0.05s-1),

t is the time in seconds.

Rearranging for t, we get:

t = (ln[A]0 - ln[A]t) / k

t = (ln(2.00) - ln(0.30)) / 0.05

t = (0.6931 - (-1.204)) / 0.05

t = 37.942 seconds (rounded to three significant figures).

Therefore, it will take approximately 37.942 seconds for the concentration of A to decrease from 2.00 moles/liter to 0.30 moles/liter.

Answer 2

34.7s elapse before the concentration of A has fallen to 0.30 M

Step-by-step explanation:

A reaction that is first order follows the equation:

ln [A] = ln [A]₀ - kt

Where [A] is concentration of A (0.30M) in time t, and [A]₀ is initial concentration (2.00M). K is reaction rate constant (0.05s⁻¹).

Replacing:

ln [0.30] = ln [2.00] - 0.05s⁻¹t

t = 34.7s elapse before the concentration of A has fallen to 0.30 M