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3 years ago

If the concentration of a reactant is 0.0560 M 25.5 seconds after a

Mathematics

1 answer:

Marysya12 [62]3 years ago

8 0

Answer:

It will take 94.54 s for the reactant concentration to decrease to 0.00450 M.

Step-by-step explanation:

A first order reaction has the general expression

(dC/dt) = -kC (Minus sign because it's a rate of reduction)

where C is the concentration of the reactant at any time

C₀ = initial concentration of the reactant

(dC/dt) = -kC

(dC/C) = -kdt

∫ (dC/C) = -k ∫ dt

Solving the two sides as definite integrals by integrating the left hand side from C₀ to C and the Right hand side from 0 to t.

We obtain

In (C/C₀) = -kt

(C/C₀) = e⁻ᵏᵗ

C(t) = C₀ e⁻ᵏᵗ

To put the final expression in a simpler format

In (C/C₀) = -kt

In C - In C₀ = -kt

In C = In C₀ - kt

Let (In C₀) = A

In C = A - kt

If the concentration of a reactant is 0.0560 M 25.5 seconds after a reaction starts and is 0.0156 M 60.5 seconds after the start of the

reaction.

C = 0.0560 M, t = 25.5 s

C = 0.0156 M, t = 60.5 s

In C = A - kt

In 0.056 = A - 25.5k

-2.8824 = A - 25.5k (eqn 1)

In 0.0156 = A - 60.5k

-4.1605 = A - 60.5k (eqn 2)

Writing the simultaneous eqns together

-2.8824 = A - 25.5k

-4.1605 = A - 60.5k

Subtract eqn 2 from eqn 1

-2.8824 + 4.1605 = -25.5k + 60.5k

1.2781 = 35k

k = (1.2781/35) = 0.03652 s⁻¹

-2.8824 = A - 25.5k

A = (25.5×0.03652) - 2.8824 = -1.9512

A = In C₀

C₀ = e^(A) = e^(-1.9512) = 0.1421 M

how many seconds after the start of the reaction does it take for the reactant concentration to decrease to 0.00450 M?

What is t when C = 0.00450 M

In C = In C₀ - kt

In 0.0045 = In 0.1421 - 0.03652t

-5.4037 = -1.9512 - 0.03652t

0.03652t = -1.9512 + 5.4037 = 3.4525

t = (3.4525/0.03652) = 94.54 s

Hope this Helps!!!

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