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.
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.
oh shoot, sorry..
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.
Sure! Let's go step by step through each problem and explain how to find the value of the equilibrium constant.
Problem 4:
In this problem, we are given the initial amounts of substances A and B before they react, and the amount of B at equilibrium. We need to calculate the value of the equilibrium constant.
1. Write the balanced chemical equation for the reaction. Without this information, we cannot determine the equilibrium constant. Please provide the equation so that we can continue with the explanation.
Problem 3:
In this problem, we are given the initial amounts of substance A2B3 and the amount of B at equilibrium. We need to find the value of the equilibrium constant.
1. Write the balanced chemical equation for the reaction. Without this information, we cannot determine the equilibrium constant. Please provide the equation so that we can continue with the explanation.