Here is a better chart for the question. Please answer all the questions.
2 ICl + H2 🡪 I2 + 2HCl
Initial Concentration (mol/L)
[ICl] [H2] Initial Rate of Formation of Products (mol/L•s)
0.10 0.10 0.0015
0.20 0.10 0.0030
0.10 0.050 0.00075
At a certain temperature the following data was collected for the reaction:
2 ICl + H2 🡪 I2 + 2HCl
Initial Concentration (mol/L)
Initial Rate of Formation of Products (mol/L•s)
[ICl][H2]
0.10
0.10
0.0015
0.20
0.10
0.0030
0.10
0.050
0.00075
Determine the order of reaction for both reactants (show your work and indicate which trials were used to determine order).
Determine k (show your work).
Write the rate law for this set of data, including the k value found in part b.
2 answers
Spacing is a real problem with html. I use periods to space. It isn't perfect but it makes it a little easier to read. I've a trial column.
Trial........[ICl]........ [H2]........... Initial Rate of Formation of Products (mol/L•s)
...1........0.10....... 0.10............... 0.0015
...2........0.20....... 0.10............... 0.0030
....3........0.10....... 0.050............. 0.00075
You determine the order by picking trial 1 and 2. You want the same concentration for one of the components. That give the order to ICl. Then you pick trial 1 and 3 which gives you the order for H2. Trial 1 and 2 are handled this way. This is hard to do on the computer but stay with me. a and be are order numbers. Divide rate 2 by rate 1
......rate = k(ICl)^a(H2)^b
.... 0.0030 = k(0.20)^a(0.10)^b
-------------------------------------------------
......0.0015 = k(0.10)^a(0.10)^b
You see (0.010)^b cancels along with k leaving
2 = 2^a which makes a = 1 so the order for ICl is 1
Do the same for the order for H2 by dividing trial 1 by trial 3.
0.0015 = k(0.10)^a (0.10)^b
------------------------------------------
0.00075 = k(0.10)^a (0.05)^b
Note that k and 0.10)^a cancels leaving
2 = 2^b which makes b = 1 so the order for H2 is 1.
Determine k:
Pick any trial and calculate k
rate = k(ICl)^1*(H2)^1
Plug in the values from the chart and solve for k.
Then you can write the rate law. You should confirm the calculations above. It's late and I could have made a typo.
Trial........[ICl]........ [H2]........... Initial Rate of Formation of Products (mol/L•s)
...1........0.10....... 0.10............... 0.0015
...2........0.20....... 0.10............... 0.0030
....3........0.10....... 0.050............. 0.00075
You determine the order by picking trial 1 and 2. You want the same concentration for one of the components. That give the order to ICl. Then you pick trial 1 and 3 which gives you the order for H2. Trial 1 and 2 are handled this way. This is hard to do on the computer but stay with me. a and be are order numbers. Divide rate 2 by rate 1
......rate = k(ICl)^a(H2)^b
.... 0.0030 = k(0.20)^a(0.10)^b
-------------------------------------------------
......0.0015 = k(0.10)^a(0.10)^b
You see (0.010)^b cancels along with k leaving
2 = 2^a which makes a = 1 so the order for ICl is 1
Do the same for the order for H2 by dividing trial 1 by trial 3.
0.0015 = k(0.10)^a (0.10)^b
------------------------------------------
0.00075 = k(0.10)^a (0.05)^b
Note that k and 0.10)^a cancels leaving
2 = 2^b which makes b = 1 so the order for H2 is 1.
Determine k:
Pick any trial and calculate k
rate = k(ICl)^1*(H2)^1
Plug in the values from the chart and solve for k.
Then you can write the rate law. You should confirm the calculations above. It's late and I could have made a typo.
0 answers