Asked by Hannah
The equilibrium constant for the reaction,
H2(g) + I2(g) == 2 HI(g) is 54.9 at 699.0 K (Kelvin).
What is the equilibrium constant for
4 HI(g) == 2 H2(g) + 2 I2(g) under the same conditions?
Note: the == indicates the equilibrium double arrow
Since the second equation is doubled I multiplied 54.9 by 2 and got 109.8. Is this correct??
H2(g) + I2(g) == 2 HI(g) is 54.9 at 699.0 K (Kelvin).
What is the equilibrium constant for
4 HI(g) == 2 H2(g) + 2 I2(g) under the same conditions?
Note: the == indicates the equilibrium double arrow
Since the second equation is doubled I multiplied 54.9 by 2 and got 109.8. Is this correct??
Answers
Answered by
bobpursley
No.
for the second equation, try this
kc=(1/54.9)^2
if you cannot figure out why, state so.
for the second equation, try this
kc=(1/54.9)^2
if you cannot figure out why, state so.
Answered by
Hannah
I do not understand why do you 1/54.9??
Answered by
DrBob222
You take 1/2 of an equation, new k = sqrt old k.
You double an equation, new k = old k^2
You revere an equation, new k = 1/old k.
You double an equation, new k = old k^2
You revere an equation, new k = 1/old k.
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