Question

Consider the reaction SO₂Cl₂(g) = SO₂(g) + Cl₂(g). The rate law for this reaction is rate = k[SO₂Cl₂] where k=2.90 ´ 10^-4 s^-1 at a particular temperature.

If the initial [SO₂Cl₂] = 0.0500 M, what is the initial rate?

What is the value of the half-life for this initial concentration of SO₂Cl₂ ?

What will be the half-life if the initial [SO₂Cl₂] = 0.0250 M

What will be the half-life if the initial [SO₂Cl₂] = 0.100 M

What will be the [SO₂Cl₂] after 2.00 minutes when we start with an initial [SO₂Cl₂] = 0.0500 M

How much time will it take for the [SO₂Cl₂] to decrease from 0.0500 M to 0.0250 M

How much time will it take for the [SO₂Cl₂] to decrease from 0.100 M to 0.0750 M

Answer 1

rate = k[SO2Cl2]

1) if [SO2Cl2] = 0.0500 M

rate = 2.90 x 10^-4 [0.0500]

rate = 1.45 x 10^-5 M/s

2) for first order reaction

t_1/2 = 0.693 / K = 0.693 / 2.90 x 10^-4

t_1/2 = 2389.65 s

3) half is same for any initial concentration

t_1/2 = 2389.65 s

4 ) t_1/2 = 2389.65 s

5) for first order

K = (2.303 / t) log [A₀] / [A]

2.90 x 10^-4 = (2.303 / 2.0 x60) log [0.05] / [A]

log [0.05] / [A] = 0.0151

-1.30 - log [A ] = 0.0151

log [A] = -1.3151

[A] = 0.0484 M

6) 2.90 x 10^-4 = (2.303 / t) log [0.05] /[0.025]

2.90 x 10^-4 = (2.303 /t) 0.301

2.303/ t = 0.0009634

t = 2390.5 s

7) 2.90 x 10^-4 = (2.303 /t) log [0.1] /[0.075]

2.90 x 10^-4 = (2.303 /t) 0.125

2.303 /t = 0.00232

t = 992.67 s