# Consider the equation A(aq) + 2B(aq) 3C(aq) + 2D(aq). In one experiment, 45.0 mL of 0.050 M A is mixed with 25.0 mL 0.100 M B. At equilibrium the concentration of C is 0.0410 M. Calculate K.

a) 7.3
b) 0.34
c) 0.040
d) 0.14
e) none of these

the answer for this is suppose to be C but i am getting .000195 can any one show me how to do it

Question ID
525257

Created
April 4, 2011 4:32pm UTC

Rating
5

2

Views
6917

1. A = (45.0 mL x 0.05M/70 mL total) = 0.03214M.
B = (25.0 mL x 0.1M/70 mL total) = 0.03571M.
C = 0
D = 0

..............A + 2B ==> 3C + 2D
initial..0.03214..0.0371..0....0
change see below.
equil....................0.0410M

change:
(C) was 0 initially, must have changed by +0.0410M.
(D) must be at equilibrium 0.0410 x (2/3) = 0.02733M so it changed by +0.02733M
A must have changed by -(0.0410/3) = -0.01367M. Just add initial + change (a negative number) to arrive at the equilibrium value.
B must have changed by -(0.0410/3)*2 = -0.02733M. Just add initial + change (a negative number) for equilibrium value.
Now plug in the numbers to the K expression. I obtained 0.03969 which rounds to 0.0397 to 3 significant figures and 0.04 looks to be the best choice. I expect we just rounded differently; you will notice i carried my numbers to four places.

525707

Created
April 5, 2011 5:19am UTC

Rating
10

2. Well, that's quite a chemical equation you've got there! Let me see if I can help you figure out the right answer.

To find the value of K, we need to use the concentrations of the reactants and products at equilibrium. From the equation, we can see that the ratio of concentrations between A, B, C, and D is 1:2:3:2.

In the given experiment, the initial concentration of A is 0.050 M and 45.0 mL (0.0450 L) of A is used. This means the initial number of moles of A is (0.050 M) * (0.0450 L) = 0.00225 moles.

Similarly, the initial number of moles of B is (0.100 M) * (0.0250 L) = 0.00250 moles.

Now, let's see how these reactants produce products at equilibrium.

Since the ratio of A to C is 1:3, the number of moles of C produced at equilibrium is 3 * (0.00225 moles) = 0.00675 moles.

Therefore, the concentration of C at equilibrium is (0.00675 moles) / (0.0450 L + 0.0250 L) = 0.0675 moles/L = 0.0675 M.

Now, to find K, we need to divide the concentration of C at equilibrium by the concentrations of A and B at equilibrium. The concentration of A is (0.00225 moles) / (0.0450 L + 0.0250 L) = 0.0225 M. The concentration of B is (0.00250 moles) / (0.0450 L + 0.0250 L) = 0.025 M.

K = (0.0675 M) / (0.0225 M * 0.025 M^2) = 1 / (0.0225 * 0.025) = 1 / 0.0005625.

And after some calculations, I get K ≈ 1777.78.

So, it seems that the correct answer is "none of these", as none of the choices provided match the value I calculated.

Keep in mind that this is just how I approached the problem, and I definitely could have made some mistakes along the way. I would recommend double-checking your calculations and making sure you're using the correct values.

Good luck!