In a first order decomposition reaction, 50% of a compound decomposes in 10.5 min. What is the rate constant of the reaction? How long does it take for 75% of the compound to decompose?

Question

In a first order decomposition reaction, 50% of a compound decomposes in 10.5 min. What is the rate constant of the reaction? How long does it take for 75% of the compound to decompose?

My professor did this problem in class:

k = .693 / 10.5 mins
= .066 min ^-1

where did .693 come from?

Thanks

Answer 1

it's from the half-life formula for first order process rate t1/2 = .693 / k

Answer 2

0.693 is the natural log of 2.

It comes from using the expression
ln(No/N) = kt
ln(100/50) = k*t_1/2
ln 2 = k*t<sub<1/2
k = ln2/t_1/2 and
k = 9.693/t_1/2

The No/N I used above as 100 and 50 can be any number choose as long as N is 1/2 of No.

Answer 3

The value 0.693 comes from the natural logarithmic constant, ln(2).

In a first-order decomposition reaction, the rate of reaction follows a first-order kinetics equation, which can be expressed as:

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

Where [A]t is the concentration of the compound at time t, [A]0 is the initial concentration, k is the rate constant, and t is the time.

If 50% of the compound decomposes in 10.5 minutes, it means that [A]t/[A]0 = 0.5. Plugging these values into the equation above, we can solve for k:

ln(0.5) = -k * 10.5 min

We can rewrite this equation as:

ln(0.5) / 10.5 min = -k

To simplify the equation further, we use the fact that ln(0.5) is equal to -ln(2), so we get:

-ln(2) / 10.5 min = -k

Therefore, k = ln(2) / 10.5 min.

The natural logarithmic constant ln(2) is approximately equal to 0.693, so we substitute this value into the equation:

k ≈ 0.693 / 10.5 min
k ≈ 0.066 min^-1

Hence, the rate constant of the reaction is approximately 0.066 min^-1.

To determine how long it takes for 75% of the compound to decompose, we can rearrange the first-order kinetics equation:

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

Substituting [A]t/[A]0 = 0.75 (since 75% of the compound decomposes) and rearranging for t:

t = -ln(0.75) / k

Plugging in the value of k (0.066 min^-1), we can calculate the time it takes for 75% of the compound to decompose.

Answer 4

The value of 0.693 that your professor used in the calculation is actually the natural logarithm of 2, denoted as ln(2). The natural logarithm of a number can be calculated using a scientific calculator or online tool.

In the context of first-order decomposition reactions, the rate constant (k) can be determined using the formula:

k = ln(2) / t(1/2)

Here, t(1/2) represents the half-life of the reaction, which is the time it takes for 50% of the compound to decompose.

In the given problem, it is stated that 50% of the compound decomposes in 10.5 minutes. Since this is the half-life of the reaction, we can substitute this value into the equation:

k = ln(2) / 10.5 min

Simplifying this expression gives:

k ≈ 0.693 / 10.5 min

So, the correct rate constant for the reaction is approximately 0.066 min^(-1).

To determine how long it takes for 75% of the compound to decompose, you can use the same formula with the updated percentage:

t = ln(3) / k

In this case, the natural logarithm of 3, denoted as ln(3), is used because we want to find the time it takes for 75% (or three halves) of the compound to decompose.

Substituting the rate constant value we calculated earlier, we get:

t ≈ 1.099 / 0.066 min^(-1)

Simplifying this expression gives:

t ≈ 16.65 minutes

Therefore, it would take approximately 16.65 minutes for 75% of the compound to decompose.