Asked by AMI
For the system 2A (g) + B( g) = 2C (g), if the initial concentration of pure C is 1.00 M, the equilibrium concentration is 0.40 M. What is the value of Kc?
Choose one answer.
a. 5.0
b. 0.45
c. 0.67
d. 2.2
e. 1.5
Choose one answer.
a. 5.0
b. 0.45
c. 0.67
d. 2.2
e. 1.5
Answers
Answered by
DrBob222
Set up an ICE chart and solve. Post your work if you get stuck.
Answered by
AMI
2A + B= 2C
I 0 0 1.00
C 2X X -2X
E 2X X 1.00-2X
KC= 0.40= C^2/ A^2*B
square root 0.40 (x*x)=1.00-2x
0.40(X)^2 SQUARE ROOT 0.40-1.00
I ended up with 0.09 but when I plugged it back in then I don't get any of the above answers. I don't know what I was doing wrong.
I 0 0 1.00
C 2X X -2X
E 2X X 1.00-2X
KC= 0.40= C^2/ A^2*B
square root 0.40 (x*x)=1.00-2x
0.40(X)^2 SQUARE ROOT 0.40-1.00
I ended up with 0.09 but when I plugged it back in then I don't get any of the above answers. I don't know what I was doing wrong.
Answered by
DrBob222
2A + B ==> 2C
If Ceq = 0.4 and it was 1.0 initially, it must have changed by 1-0.4= 0.6
Therefore, B must have changed by +0.3 and A must hav changed by 0.6 so the final concns are
A = 0.6 M
B = 0.3 M
C = 0.4 M
Plug those in and calculate Kc.
If Ceq = 0.4 and it was 1.0 initially, it must have changed by 1-0.4= 0.6
Therefore, B must have changed by +0.3 and A must hav changed by 0.6 so the final concns are
A = 0.6 M
B = 0.3 M
C = 0.4 M
Plug those in and calculate Kc.
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