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!!!
