A2B3 --> 2A + 3B
I 0.001 0 0
C –x 2x 3x
E 0.001-x 2x 3x
3x = 0.006, x = 0.002
A = 0.004
B = 0.006
A2B3 = 0.001-0.002 =
(0.004)2(0.006)3/
Am I close?
I think we agree on A and B but not A2B3. I answered it below. A2B3 = 0.01-0.002 = 0.008
Don't forget that you must convert mols to M before calculating Kc.
3x = 0.006, x = 0.002
A = 0.004
B = 0.006
A2B3 = 0.01-0.002 = 0.008
(0.004)2(0.006)3/(0.008) = 4.32x10-10
No. I didn't calculate it but I know it isn't right becasue you didn't convert mols to M. M = mols/L so you must divide each of the mols by 0.200 L to convert to M. M is what goes into K expression. I thinkl 200 mL is correct although it isn't posted in this post.
Yes, you are close. But there is a small error in your calculation for the value of A2B3. Let's go through the steps again to correct it.
First, let's determine the values of A and B using the given information:
- We are given that 3x = 0.006 and x = 0.002, so we can calculate x = 0.002.
- Since A is produced by consuming x, the value of A can be calculated as A = 0.001 - x = 0.001 - 0.002 = -0.001.
- Similarly, B is produced by consuming 2x, so the value of B can be calculated as B = 2x = 2(0.002) = 0.004.
Now, let's calculate the value of A2B3:
- A2B3 is formed by combining 2 moles of A and 3 moles of B. Given that A = -0.001 and B = 0.004, we can calculate A2B3 as (A)2(B)3 = (-0.001)2(0.004)3 = 0.000008.
So, the correct value of A2B3 is 0.000008.
In summary, the correct calculations are:
A = 0.001 - x = 0.001 - 0.002 = -0.001
B = 2x = 2(0.002) = 0.004
A2B3 = (A)2(B)3 = (-0.001)2(0.004)3 = 0.000008.