please help!
If the initial pressure of I2(g) is 1.738 atm, calculate the % decomposition of I2(g) when the reaction comes to equilibrium according to the balanced equation. The value of Kp at 1000 K is 0.254. The initial pressure of the reaction products is 0 atm.
I2(g) = 2I(g)
7 answers
Set up an ICE chart, substitute into Kp expression, and solve.
i did.. it was wrong for the last time
for the last time i did it*
If you post what you did perhaps I can find the error.
initial:
I2 = 1.738 atm
2I = 0
change:
I2 = -2x
2I = x
equilibrium:
I2 = 1.738 - 2x
2I = x
so i set the equation as
0.254 = x/(1.738 - 2x)
and i solved for x. i got 0.292
and after that how do i get % decomposition?
I2 = 1.738 atm
2I = 0
change:
I2 = -2x
2I = x
equilibrium:
I2 = 1.738 - 2x
2I = x
so i set the equation as
0.254 = x/(1.738 - 2x)
and i solved for x. i got 0.292
and after that how do i get % decomposition?
equilibrium:
I^- (not 2I) = 2x
I2 = 1.738-x
Then Kp = pI^-^2/pI2
0.254 = (2x)^2/(1.738-x).
Solve for x. You didn't have 2x and then didn't square it.
%decomposition = (pI^-/1.738)*100 = ??
I^- (not 2I) = 2x
I2 = 1.738-x
Then Kp = pI^-^2/pI2
0.254 = (2x)^2/(1.738-x).
Solve for x. You didn't have 2x and then didn't square it.
%decomposition = (pI^-/1.738)*100 = ??
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