The decomposition of HCO2H follows first-order kinetics.

HCO2H(g) CO2(g) + H2(g)

The half-life for the reaction at a certain temperature is 44 seconds. How many seconds are needed for the formic acid concentration to decrease by 73%?

k = 0.693/t1/2

ln(No/N) = kt
No = any convenient number but I would use 100 to represent 100%. Then if it decreases by 73% that leaves 27% so
N = 27
k from above
t = ? Solve for t in seconds if you use half life in seconds.

Thank you! You are legendary!

To determine the time required for the formic acid concentration to decrease by 73%, we can use the equation for first-order kinetics:

ln([A]/[A]0) = -kt

Where:
[A] = concentration at time t
[A]0 = initial concentration
k = rate constant of the reaction
t = time

In this case, we want to find the time required for the concentration to decrease by 73%. So, [A]/[A]0 = 0.27 (73% decrease means 27% remaining).

ln(0.27) = -k(44s)

Let's solve for k first. Rearranging the equation:

k = -ln(0.27)/44s

Now we can find the time needed for the concentration to decrease by 73%. Let's call it t_needed.

ln([A]/[A]0) = -k * t_needed
ln(0.27) = -k * t_needed

Substituting the value of k:

ln(0.27) = -(-ln(0.27)/44s) * t_needed

Let's solve for t_needed:

t_needed = ln(0.27) * 44s / ln(0.27)

Calculating this value, we get:

t_needed ≈ 121.33 seconds

Therefore, it would take approximately 121.33 seconds for the formic acid concentration to decrease by 73%.

To determine the time needed for the formic acid concentration to decrease by 73%, we can use the concept of half-life.

In a first-order reaction, the half-life is the time required for the concentration of the reactant to decrease by half. Since we are given the half-life of the reaction, we can use it to find the rate constant (k) and then determine the time needed for a specific percentage decrease.

The half-life of a first-order reaction can be calculated using the equation:

t1/2 = (0.693 / k)

Given that the half-life (t1/2) is 44 seconds, we can rearrange this equation to solve for the rate constant (k):

k = 0.693 / t1/2

k = 0.693 / 44

k ≈ 0.0157 s^(-1)

With the rate constant determined, we can now calculate the time required for a 73% decrease in concentration.

First, we need to find the number of half-lives required for a 73% decrease. Since each half-life corresponds to a 50% decrease, the number of half-lives can be calculated using the equation:

n = log(73/100) / log(1/2)

n ≈ 1.359

The number of half-lives rounded up to 2 since we cannot have a fraction of a half-life.

Now, to find the time required for a 73% decrease, we multiply the number of half-lives (2) by the given half-life (44 seconds):

time required = number of half-lives * half-life

time required = 2 * 44

time required = 88 seconds

Therefore, it would take approximately 88 seconds for the formic acid concentration to decrease by 73%.