An experiment to study a reaction A P gave the following data. Linear regression

Question

An experiment to study a reaction A P gave the following data. Linear regression analysis of the linear graph gave the equation y = mx + b with b = 4.57688 and m = 5.16677. 0.25 0.20 0.15 0.10 0.05 0.00 0.22 0.0674 0.0395 0.0280 0.0218 0.0178 60 50 40 30 20 10 4 0 2 4 6 8 10 6 time (s) 10 2.0 2.5 4 3.0 3.5 4.0 0 246 8 10 time (s) 0 2 4 6 810 time (s) Part A. Determine the order of the reaction. (be sure you can justify your answer in a short sentence in the real exam) Part B. Determine the rate constant Part C. Determine the half-life. (be sure you can show all your work) Part D, and E Determine the concentrations of A and P at t= 3s Part F. Determine the rate of the reaction at t= 3s

Explanation / Answer

The best way to identify fast and easy the rate of reaction AND the rate constant is via Graphical Method.

First, as the name implies, we need to graph all types of order (most common)

Zero = C vs t;

For zero order, there is no dependency of concentrations:

dC/dt = k*C^0

dC/dt = k

When developed:

C = C0 - kt

if x axis is "time" then the slope is "k", and y-intercept is initial concentration C0. y-axis if C (concentration)

First = ln(C) vs. t

For first order

dC/dt = k*C^1

dC/dt = k*C

When developed:

dC/C = k*dt

ln(C) = ln(C0) - kt

if x axis is "time" then the slope is "-k", and y-intercept is initial concentration C0. y-axis if ln(C) (natural logarithm of concentration)

Second = 1/C vs. t

For Second order

dC/dt = k*C^2

When developed:

dC/C^2 = k*dt

1/C= 1/C0 + kt

if x axis is "time" then the slope is "k", and y-intercept is initial concentration C0. y-axis if 1/(C (inverse of concentration)

Know, graph all data in the 3 graphical methods

then

from the best fit

this is 1/A vs. t

which is 2nd order

b)

rate constnat --> slope

slope -> m = 5.16677

k = 5.16677 1/(M-s)

c)

half life:

HL = 1/(k*[A]0)

HL = 1/(5.16677 *0.22) = 0.87974 seconds

d+e)

concnetration of A and P at t = 3s

1/A = 1/A0 + kt

1/A = 1/0.22 + 5.16677 *3

A = ( 1/0.22 + 5.16677 *3)^-1= 0.04988 M

if this is true

then

P= A0 - Aleft = (0.22 -0.04988) = 0.17012 M

f.

rate ate t = 3s

Rate = k*[A]^2

Rate = (5.16677)(0.04988 ^2) = 0.012854 M/s

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