need help with questions related to rate. i pasted the two links below. any help would be appreciated!!
gyazo.com/425bad92c5dfaae93a8e484ffd9ae8ca
gyazo.com/5a0190f5c6ccec1a96263cb2342fad2e
6 answers
for the second link, i figured out a-c, just need help with d. is it just finding the slope of the time/concentration above and below 493?
I'll get you started. I can't copy the table here. First you want to find the exponents for X and yY The exponents are the orders for X and Y. For example, rate = k(X)^a(Y)^b. a is the order for X and b is the order for b. To do this, look at the chart. You want to find a trial that uses the same concn of either but with a different concn for the other. I picked trials 1 and 2. You see Y stays the same but X is different. Now write the rate equations.
rate 1 = k(X)^a(Y)^b, then I'll divide the data for 2 by the date from 1 like this
rate 2 = k(X)^a*(Y)^b) = 1.025 = k(0.3)^a*(0.5)^b
----------------------------------------------------------------
rate 1 = k(X)^a*(Y)^b = 0.6830 = k(0.2)^a*(0.5)^b
That -------------- is a division line.
That leaves us with
1.025 = k(0.3)^a*(0.5)^b
----------------------------------- = 1.5 = k(1.5)^a which makes a = 1
0.6830 = k(0.2)^a*(0.5)^b
therefore, the order for x is 1.
Do the same for experiments 1 and 3 to find exponent b which is the order for Y.
Then substitute numbers from any trial with what you know now to be the rate law to find k. From the complete law of
rate = k(X)(Y). Plug in the units for X, Y, and rate and solve for units for k. I didn't look at the second link.
Post your work if you get stuck. If you have questions please BE SPECIFIC about what you don't understand.
T
rate 1 = k(X)^a(Y)^b, then I'll divide the data for 2 by the date from 1 like this
rate 2 = k(X)^a*(Y)^b) = 1.025 = k(0.3)^a*(0.5)^b
----------------------------------------------------------------
rate 1 = k(X)^a*(Y)^b = 0.6830 = k(0.2)^a*(0.5)^b
That -------------- is a division line.
That leaves us with
1.025 = k(0.3)^a*(0.5)^b
----------------------------------- = 1.5 = k(1.5)^a which makes a = 1
0.6830 = k(0.2)^a*(0.5)^b
therefore, the order for x is 1.
Do the same for experiments 1 and 3 to find exponent b which is the order for Y.
Then substitute numbers from any trial with what you know now to be the rate law to find k. From the complete law of
rate = k(X)(Y). Plug in the units for X, Y, and rate and solve for units for k. I didn't look at the second link.
Post your work if you get stuck. If you have questions please BE SPECIFIC about what you don't understand.
T
how do you figure out the power of mol, L, and s?
thank you for your help!
From my first response, look at lines 5 and 4. I have copied them here.
From the complete law of
rate = k(X)(Y)^b and I calculated b to be 1 also Under the initial rate column of the chart you see that has units of mols/L*s Plug that in for rate. Look at the chart and concn for X and Y are in mols/L (or M). Plug mols/L in for units of X and Y. If you solve the equation first for k you get
k = initial rate/(X)(Y). Pluging in initial rate of mols/L*s and mols/L for X and mols L for Y, solve for units for k. That should do it.
From the complete law of
rate = k(X)(Y)^b and I calculated b to be 1 also Under the initial rate column of the chart you see that has units of mols/L*s Plug that in for rate. Look at the chart and concn for X and Y are in mols/L (or M). Plug mols/L in for units of X and Y. If you solve the equation first for k you get
k = initial rate/(X)(Y). Pluging in initial rate of mols/L*s and mols/L for X and mols L for Y, solve for units for k. That should do it.
Here is a link for the Wikipedia article on rate laws.
https://en.wikipedia.org/wiki/Reaction_rate_constant
Look under section 4 and you see this ( but it might help to read the entire article).
For order (m + n), the rate constant has units of mol1−(m+n)·L(m+n)−1·s−1
For order zero, the rate constant has units of mol·L−1·s−1 (or M·s−1)
For order one, the rate constant has units of s−1
For order two, the rate constant has units of L·mol−1·s−1 (or M−1·s−1)
And for order three, the rate constant has units of L2·mol−2·s−1 (or M−2·s−1)
https://en.wikipedia.org/wiki/Reaction_rate_constant
Look under section 4 and you see this ( but it might help to read the entire article).
For order (m + n), the rate constant has units of mol1−(m+n)·L(m+n)−1·s−1
For order zero, the rate constant has units of mol·L−1·s−1 (or M·s−1)
For order one, the rate constant has units of s−1
For order two, the rate constant has units of L·mol−1·s−1 (or M−1·s−1)
And for order three, the rate constant has units of L2·mol−2·s−1 (or M−2·s−1)
0 answers