thank you 10. The equilibrium system 2A2B Initially 3.0 moles of A are placed in

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10. The equilibrium system 2A2B Initially 3.0 moles of A are placed in a 1.5-L flask. Determine the concentration of C at equilibrium. C has a very small equilibrium constant: K 2.6 x 10 11, A general reaction written as A + 2B C + 2D is studied and yields the following data: [Alo 0.150 M 0.150 M 0.300 M [Blo Initial [CIA! 0.150 M 8.00 × 10-3 mol/Ls 0.300 M 1.60 × 10-2 mol/Ls 0.150 M 3.20 x 10 mol/L's a. Determine the rate law for the process. What is the overall order of the reaction? b. What is the numerical value of the rate constant? c. Determine the initial rate of B consumption (A[BA) for the first trial? d. Determine the initial rate of C production (AlC A) irlA) 0.200 Mand [Bl-0.500 M. and [B]-0.500 M

Explanation / Answer

For problem 10:

We make our BCA table:

K = 2.6 x 10-6 = [B]2[C] / [A]2

2.6 x 10-6 = [2x]2[x] / [3-2x]2

2.6 x 10-6 = [4x3] / [3-2x]2

Solving for x:

x = 0.0179 moles

[C] = 0.0179 moles / 1.5L = 0.01193M

For problem 11:

a) We use ratios of rates to get the exponents of rate law. It follows:

rate = k*[A]m*[B]n

To obtain m:

rate 1/ rate 3 = k*[0.15]m[0.15]n / k*[0.3]m[0.15]n

0.008 / 0.032 = 0.15m/0.3m

0.25 = 0.5m

ln 0.25 / ln 0.5 = m

m = 2

To obtain n:

rate 1/ rate 2 = k*[0.15]m[0.15]n / k*[0.15]m[0.3]n

0.008 / 0.016 = 0.15n/0.3n

0.5 = 0.5n

n = 1

So, rate law:

rate = k[A]2[B]

b) For this letter, we can insert any data point in the obtained law:

0.008 = k * [0.15]2[0.15]

k = 0.008 / 0.153

k = 0.008 / 0.003375 = 2.37 1/mol2-L-s

c) For obtaining rate of B, we use stoichiometric relationship:

rb/2 = rc/1

rb = 2rc = 2 * 0.008 = 0.016 mol/L-s

d) We insert values in obtained law:

rate = 2.37 * [0.2]2 [0.5] = 0.0474 mol/L-s

2A <-> 2B + C B 3 moles 0 0 C -2x +2x x A 3-2x 2x x

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