# Problem: The reaction2SO3(g) → 2SO2(g) + O2(g)is second order. If the concentration of SO3 decreases from 0.0360 moles/L to 0.0075 moles/L in 863 s, what is the rate constant for the reaction?a) 0.476 moles-1 L  s−1b) 0.0987 moles-1 L  s−1c) 5.23 moles-1 L  s−1d) 0.376 moles-1 L  s−1e) 0.122 moles-1 L  s−1

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81% (209 ratings)
###### FREE Expert Solution

We’re being asked to determine the rate constant (k) of a second-order reaction:

2 SO3(g)  2 SO2(g) + O2(g)

The concentration of SO3 decreases from 0.0360 moles/L to 0.0075 moles/L in 863 s.

The integrated rate law for a second-order reaction is as follows:

$\overline{)\frac{\mathbf{1}}{{\mathbf{\left[}\mathbf{A}\mathbf{\right]}}_{\mathbf{t}}}{\mathbf{=}}{\mathbf{kt}}{\mathbf{+}}\frac{\mathbf{1}}{{\mathbf{\left[}\mathbf{A}\mathbf{\right]}}_{\mathbf{0}}}}$

where:

[A]t = concentration at time t

k = rate constant

t = time

[A]0 = initial concentration

81% (209 ratings) ###### Problem Details

The reaction

2SO3(g) → 2SO2(g) + O2(g)

is second order. If the concentration of SO3 decreases from 0.0360 moles/L to 0.0075 moles/L in 863 s, what is the rate constant for the reaction?

a) 0.476 moles-1 L  s−1

b) 0.0987 moles-1 L  s−1

c) 5.23 moles-1 L  s−1

d) 0.376 moles-1 L  s−1

e) 0.122 moles-1 L  s−1

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Integrated Rate Law concept. You can view video lessons to learn Integrated Rate Law. Or if you need more Integrated Rate Law practice, you can also practice Integrated Rate Law practice problems.

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Based on our data, we think this problem is relevant for Professor Bindell's class at UCF.