The light-stimulated conversion of 11-cis-retinal to 11-trans-retinal is central to the vision process in humans. This reaction also occurs (more slowly) in the absence of light. At 80.0 ∘C in heptane solution, the reaction is first order with a rate constant of 1.02×10−5/s.

How many hours does it take for the concentration of 11-trans-retinal to reach 3.14×10−3 M ? (Note: 11-cis-retinal + 11-trans-retinal = total amount of retinal in eye)

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pstnonsonjoku pstnonsonjoku · Answer 1 · 2021-03-26T05:21:04+01:00

Answer:

See explanation

Explanation:

For first order reaction;

ln[A] = ln [A]o - kt

Where;

[A] = concentration at time = t

[A]o = initial concentration

k = rate constant

t = time taken

i) ln[A] = ln( 3.2×10−3 M) - (1.02×10−5/s * 6 * 60 * 60)

ln[A] = -5.745 - 0.22

[A] = e^-5.965

[A] = 2.6 * 10^-3

ii) Now, Amount reacted = 0.25 * 3.2×10−3 M = 8 * 10^-4 M

[A] = 3.2×10−3 M - 8 * 10^-4 M

[A] = 2.4 ×10−3 M

ln[A] = ln [A]o - kt

ln[A] - ln [A]o = - kt

t = ln[A] - ln [A]o/-k

t = ln(2.4 ×10−3 M) -ln (3.2×10−3 M)/-(1.02×10−5/s)

t= -6.032 - (-5.745)/-(1.02×10−5/s)

t = -0.287/-(1.02×10−5/s)

t = 468.95 minutes

iii) t = ln[A] - ln [A]o/-k

t = ln(3.15×10−3 M) - ln (3.2×10−3 M)/-(1.02×10−5/s)

t = -5.760 - (-5.745)/-(1.02×10−5/s)

t = -0.015/-(1.02×10−5/s)

t = 0.41 hours