calculate the specific rate constant
A] (mol/L) [B] (mol/L) Rate of appearance of C (mol/L-hr)
1.0 ×10^-2 1.0 0.30×10^−6
1.0 ×10^−2 3.0 8.10×10^−6
2.0 ×10^−2 3.0 3.24×10^−5
2.0 ×10^−2 1.0 1.20×10^−6
3.0 ×10^−2 3.0 7.30×10^−5
Do you have a question. This is merely data.
calculate the specific rate constant
between substances A and B has been found to given data
3A + 2B ---> 2C + D
You piggy backed this question onto a response to an earlier question answered by Bob Pursley. I answered this question there.
To calculate the specific rate constant, we can use the rate equation:
Rate = k[A]^m[B]^n
Where:
Rate = rate of appearance of C (mol/L-hr)
k = specific rate constant (1st order)
[A] and [B] = initial concentrations of reactants A and B (mol/L)
m and n = reaction orders with respect to reactants A and B, respectively
Looking at the given data, we can select two sets of experiments that have the same [B] value. Let's choose the first and third rows:
For the first row:
[A] = 1.0 × 10^-2 mol/L
[B] = 1.0 mol/L
Rate = 0.30 × 10^-6 mol/(L-hr)
For the third row:
[A] = 2.0 × 10^-2 mol/L
[B] = 3.0 mol/L
Rate = 3.24 × 10^-5 mol/(L-hr)
Now, let's plug these values into the rate equation:
0.30 × 10^-6 mol/(L-hr) = k(1.0 × 10^-2 mol/L)^m(1.0 mol/L)^n (equation 1)
3.24 × 10^-5 mol/(L-hr) = k(2.0 × 10^-2 mol/L)^m(3.0 mol/L)^n (equation 2)
Since [B] is constant, the ratio of the rates is equal to the ratio of the concentrations of A raised to the powers m in equation 1 and equation 2:
(0.30 × 10^-6 mol/(L-hr))/(3.24 × 10^-5 mol/(L-hr)) = (1.0 × 10^-2 mol/L)^m/(2.0 × 10^-2 mol/L)^m
Simplifying, we get:
0.30 × 10^-6/3.24 × 10^-5 = (1/2)^m
Now, solve for m:
m = log(0.30 × 10^-6/3.24 × 10^-5) / log(1/2)
Using a scientific calculator, calculate the value of m.
Once you have obtained the value of m, substitute it back into either equation 1 or equation 2 to solve for the specific rate constant k.