the rate constant for this first-order reaction is 0.0457 s^-1 at 300 °C.

Question

the rate constant for this first-order reaction is 0.0457 s^-1 at 300 °C.

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if the initial mass of A is 16.15 g, calculate the mass of A remaining after 0.730 min.

Answer 1

We can use the first-order rate law equation to determine the mass of A remaining after a certain amount of time:

ln([A]t/[A]0) = -kt (remember that ln(x/y)= ln x - ln y)

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

We can rearrange this equation to solve for [A]t:

[A]t = [A]0 e^(-kt) The problem here is that to calculate concentrations and we need to know what the MW of A is... I don't know what form the final answer should be so we're going to make some assumptions.

First, we need to convert the mass of A to its concentration using its molar mass and the volume of the reaction mixture. Let's assume a volume of 1 L for simplicity:

molar mass of A = 100 g/mol (assumed) moles of A = 16.15 g / 100 g/mol = 0.1615 mol concentration of A = moles of A / volume = 0.1615 mol / 1 L = 0.1615 M

Now we can use the rate constant and time to calculate [A]t:

k = 0.0457 s^-1 t = 0.730 min = 43.8 s

[A]t = [A]0 e^(-kt) [A]t = 0.1615 M e^(-0.0457 s^-1 × 43.8 s) [A]t = 0.1615 M e^(-2.001) [A]t = 0.1615 M × 0.1357 [A]t = 0.0219 M



Finally, we can convert the concentration back to mass using the molar mass:

mass of A remaining = [A]t × volume × molar mass of A mass of A remaining = 0.0219 M × 1 L × 100 g/mol mass of A remaining = 2.19 g

Therefore, the mass of A remaining after 0.730 min is 2.19 g.