Question

a spectrophotometry experiment was conducted for the following reaction, which occurred in a 1.10 cm cuvette. The initial concentration of the blue reaciton was 0.75A(aq) (colorless)+B(aq) (blue)→C(aq) (colorless)+D(aq) (colorless)Absorbance, A Time, t 0.77 0.00s 0.67 30.00s 0.60 60.00sa. Find the molar absorptivity coefficient, a, for the blue reactant.b. Find the initial rate of the reaction.

Answer 1

Answer:

a. 0.933 M⁻¹.cm⁻¹

b. 0.003 M/s

Explanation:

a. The molar absorptivity coefficient (ε) is the ability of 1 mol of the substance to mitigate the light of a wavelength. It means that it's a measure of how strong a substance can absorb radiation. It can be calculated by the absorbance (A), the concentration (c) and the length of the cuvette (l).

ε = A/(c*l)

Thus, at t = 0, for the blue reactant, A = 0.77, c = 0.75 M, l = 1.10 cm.

ε = 0.77/(0.75*1.10)

ε = 0.933 M⁻¹.cm⁻¹

b) The rate of the reaction can be found by the rate of desapering of the blue reactant. Let's find the concnetration when t = 30.00 s:

ε = A/(c*l)

0.933 = 0.67/(c*1.10)

1.10c = 0.67/0.933

1.10c = 0.7181

c = 0.65 M

Thus, the rate is the variation of the concentration divided by the time variation:

r = (0.75 - 0.65)/30

r = 0.003 M/s

Because the coefficient of the blue reactant is 1, the rate of reaction is equal to the rate of disappearing of B, 0.003 M/s.

Answer 2

a. The molar absorptivity coefficient is 0.933 $\rm M^-^1\;cm^-^1$. b. The initial rate of reaction is 0.003 M/sec.

a. Molar absorptivity can be defined as the absorption of 1 molar substance in the 1 cm path length.

The molar absorptivity ($\rm \epsilon$) can be given by:

$\rm \epsilon\;=\dfrac{Absorbance}{concentration\;\times\;path\;length}$

It is given that, at time 0s, the absorbance of the blue reactant = 0.77 A

concentration = 0.75 M

path length = 1.10 cm

$\rm \epsilon$ = $\rm \dfrac{0.77}{0.75\;\times\;1.10}$

$\rm \epsilon$ = 0.933 $\rm M^-^1\;cm^-^1$.

The molar absorptivity coefficient of the blue reactant is 0.933 $\rm M^-^1\;cm^-^1$.

b. The initial rate of reaction has been equal to the rate of disappearing of the blue reactant.

The rate of disappearing of the blue reactant can be given by:

$\rm \dfrac{c_1\;-\;c_2}{t}$

C1 is the concentration at time t1.

C2 is the concentration at time t2.

t is the difference in time t1 and t2.

Let c1 be the concentration at time 0s.

c1 = 0.75 M

C2 be the concentration at time 30 s.

Concentration at 30 sec, can be given by:

Absorbance = 0.67

path length = 1.10 cm

molar absorptivity coefficient = 0.933 $\rm M^-^1\;cm^-^1$.

c = $\rm \dfrac{absorbance}{\epsilon\;\times\;path\;length}$

c = $\rm \dfrac{0.67}{0.933\;\times\;1.10}$ M

c = 0.65 M

rate of disappearing of the blue reactant = $\rm \dfrac{0.75\;-\;0.65}{30}$

rate of disappearing of the blue reactant = 0.003 M/sec

The initial rate of reaction is given by 0.003 M/sec.

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