The rate constant for this first-order reaction is 0.0235 s−1 at 300 °C.

Question

The rate constant for this first-order reaction is 0.0235 s−1 at 300 °C.

A⟶products

If the initial mass of A is 19.02 g, calculate the mass of A remaining after 1.41 min.

mass of A: g

Answer 1

For a first order reaction, we can use the fact that the fraction remaining after decay is equal to 0.5ⁿ where n is the number of half lives that have elapsed during the stated time of the decay. So we'll need to know the half life (t_1/2) and again, for a first order reaction this is simply t_1/2 = ln 2 / k

For a first order reaction, the half life (t_1/2) = ln 2 / k = 0.693 / k

t_1/2 = 0.693 / 0.0235 s⁻¹

t_1/2 = 29.5 seconds

Also, for a first order reaction, fraction remaining (FR) = 0.5ⁿ where n is the number of half lives elapsed.

number of half lives elapsed = 1.41 min x 60 sec / min = 84.6 sec x 1 half life/ 29.5 sec = 2.87 half lives

FR = 0.5^2.87

FR = 0.137

Mass remaining = 0.137 x 19.02 g = 2.60 g (3 sig. figs.)

Answer 2

In order to solve this problem, we will need to use two equations. One of these is our half-life equation, and the other is an equation to solve for the final amount of product.

Half-life equation: ln(1/2) = -K x T_1/2

Final Mass EQ: M_final = M_initial x (1/2)^{T/ T1/2}

What information do we know from the problem?

1. Ist order Reaction - only the mass or concentration of 1 compound determines the rate of the EQ.
2. K = 0.0235 / s
3. T = 300 C
4. M_{0 (initial)} = 19.02 grams
5. M_{f (final)} = ??
6. t = 1.41 minutes

First, we need to convert our time into seconds 1.41 min x 60 sec/ min = 84.6 seconds

Next, we plug in our values to solve for the half life

ln (1/2) = -K x t_1/2

ln (1/2) = -0.0235/s x t_1/2 solve for ln(1/2)

-0.693 = -0.0235/s x t_1/2 divide both sides by -0.0235

29.49 s = t_1/2 Now that we have the t_1/2 we can use that in our next equation

M_f = M_o x (1/2)^t/t1/2

M_f = 19.02g x (1/2)^{84.6/ 29.49} Do exponents first: 84.6/29.49 = 2.86

(1/2)^2.86 = 0.1349

M_f = 19.02g x 0.1349

M_f = 2.56 grams This is the # of grams remaining after 84.6 seconds

Hope this helps!