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, how many seconds after the start of the reaction does it
take for the reactant concentration to decrease to 0.00450 M?
Assume the reaction is first order in this reactant.

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!!!

250

step-by-step explanation:

x=6 and x=3

beep beep imma sheep