# Problem: For the reaction given, the FeCl3 changes from 1.000 M to 0.444 M in the first 10 seconds.Calculate the average rate in the first 10 seconds. 2 FeCl3 (s) → 2 Fe (s) + 3 Cl2 (g) a. 0.0384 M/sb. 0.210 M/sc. 0.0278 M/sd. 0.0874 M/se. 0.568 M/s

###### FREE Expert Solution
85% (229 ratings)
###### FREE Expert Solution

We’re being asked to calculate the average rate of the reaction for the following reaction:

2 FeCl3(s)  2 Fe(s) + 3 Cl2(g)

For the reaction, the concentration FeCl3 changes from 1.000 M to 0.444 M in the first 10 seconds.

Recall that for a reaction aA  bB, the rate of a reaction is given by:

$\overline{){\mathbf{Rate}}{\mathbf{=}}{\mathbf{-}}\frac{\mathbf{1}}{\mathbf{a}}\frac{\mathbf{\Delta }\mathbf{\left[}\mathbf{A}\mathbf{\right]}}{\mathbf{\Delta t}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{b}}\frac{\mathbf{\Delta }\mathbf{\left[}\mathbf{B}\mathbf{\right]}}{\mathbf{\Delta t}}}$

where:

Δ[A] = change in concentration of reactants or products (in mol/L or M), [A]final – [A]initial

Δt = change in time, tfinal – tinitial

85% (229 ratings)
###### Problem Details

For the reaction given, the FeCl3 changes from 1.000 M to 0.444 M in the first 10 seconds.

Calculate the average rate in the first 10 seconds.

2 FeCl3 (s) → 2 Fe (s) + 3 Cl2 (g)

a. 0.0384 M/s

b. 0.210 M/s

c. 0.0278 M/s

d. 0.0874 M/s

e. 0.568 M/s

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 Average Rate of Reaction concept. You can view video lessons to learn Average Rate of Reaction. Or if you need more Average Rate of Reaction practice, you can also practice Average Rate of Reaction practice problems.

What is the difficulty of this problem?

Our tutors rated the difficulty ofFor the reaction given, the FeCl3 changes from 1.000 M to 0....as medium difficulty.

How long does this problem take to solve?

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What professor is this problem relevant for?

Based on our data, we think this problem is relevant for Professor Young's class at ST THOMAS.