Asked by TP
Using your knowledge of equilibrium constants, calculate the value of K1 given the following equations:
2A+B ---> C K1=x
2B ----> D K2=5.00
2C -----> 4A+D K=0.0588
The arrows are reversible
2A+B ---> C K1=x
2B ----> D K2=5.00
2C -----> 4A+D K=0.0588
The arrows are reversible
Answers
Answered by
DrBob222
(1) 2A + B ---> C K1=x
(2) 2B ==> D k2
(3) 2C --> 4A + D k3
Reverse (2) D --> 2B k = 1/k2
add to (3) 2C ==> 4A + + D k3
_________________
sum.(4).2C ==> 4A + 2B k = (1/k2)*k3 or k3/k2
Reverse (4) 4A + 2B -> 2C k = 1/k3/k2 or k2/k3 but that is just twice equation (1) so k1 = √k2/k3
Post your work if you get stuck.
(2) 2B ==> D k2
(3) 2C --> 4A + D k3
Reverse (2) D --> 2B k = 1/k2
add to (3) 2C ==> 4A + + D k3
_________________
sum.(4).2C ==> 4A + 2B k = (1/k2)*k3 or k3/k2
Reverse (4) 4A + 2B -> 2C k = 1/k3/k2 or k2/k3 but that is just twice equation (1) so k1 = √k2/k3
Post your work if you get stuck.
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