PLEASE fully explain these...I've already tried to figure them out and failed terribly... thanks so much
4. 5.6 x 10-6 mol of A and 5 x 10-5 mol of B are mixed in a 200 mL flask. The system is represented by the equation:
At equilibrium, there is 4.8 x 10-5 mol of B. Calculate the value of the equilibrium constant.
3. A 100 mL container is filled with 0.01 mol of A2B3. At equilibrium, it also contains 0.006 mol of B. Find the value of K.
4 answers
You didn't include the equation for #4.
#5 isn't all that clear but I assume that A2B3 dissociates according to the equation
..........A2B3 ==> 2A + 3B
I.........0.01......0.....0
C..........-x.......2x.....3x
E........0.01-x....2x......3x
The problem tells you that 3x = 0.006; therefore, x = 0.002
So at equilibrium A is 0.004 and A2B3 = 0.01-0.002.
Calculate M of each (mols/0.1L = ?) substitute into K expression and solve for Kc.
(Note: The confusing part to me is that "it ALSO contains 0.006 mol B. Does that mean at equilibrium it contains 0.01 mol A2B3 + 0.006 mol B? I doln't think it means that but that's what the problem says.
..........A2B3 ==> 2A + 3B
I.........0.01......0.....0
C..........-x.......2x.....3x
E........0.01-x....2x......3x
The problem tells you that 3x = 0.006; therefore, x = 0.002
So at equilibrium A is 0.004 and A2B3 = 0.01-0.002.
Calculate M of each (mols/0.1L = ?) substitute into K expression and solve for Kc.
(Note: The confusing part to me is that "it ALSO contains 0.006 mol B. Does that mean at equilibrium it contains 0.01 mol A2B3 + 0.006 mol B? I doln't think it means that but that's what the problem says.
oh shoot, sorry..
4. 2A + B --> 3C
3. A2B3 --> 2A + 3B
4. 2A + B --> 3C
3. A2B3 --> 2A + 3B
OK. I guessed right on #4. #3 is done the same way isn't it. Just plug in the numbers.
1 answer