For the reaction: H2+Br2 ⇌ 2HBr, Kc = 7.5x10^2 at a certain temperature. 1.00 mole HBr is placed in 5.0 L flask at a certain temperature. What is the concentration of HBr at equilibrium?
a) 0.19
b) 0.81
c) 0.01
d) 0.94
e) 0.03
What I got so far for the molarity,
H2 + Br2 ⇌ 2HBr
I 0 0 .2
C +x +x -2x (due to 2 moles)
E x x .2-x
-----------------------------------------------
for kc 7,5x10^2 = 750, put it over 1 due to being reverse reaction, so i get 0.0013, now plug it all in.
0.0013=x^2/(.2-2x) ----> = , (x^2) - (2.6x10^-4) + (.00266) = (0), so i plugged that into the quadratic formula and got .01485, I then did .2 - 2(.014) (as the equation said in the beginning) and got .1703, which is not an option, so any help would be kindly appreciated, and as to where I went wrong, I don't know. Thank you so much :)
Kc = 7.5*10^2 = 750
initially
[HBr] = 1/5 = 0.20
NOTE: no need to use BACKWARDS reaction... and if you do that, then you need to change the notation in "Change" in the ICE table since now, products are reactants and backwards
so...
initially
[H2] = 0
[Br2] = 0
[HBr] = 0.2
in equilibirum:
[H2] = 0 - x
[Br2] = 0 - x
[HBr] = 0.2 + 2x
NOTE: Typically, we assume forward direction, anywas, avoid inverting/reversing K and products/Reactatns, simply use maths
So..
substitute in Kc expression:
750 = (0.2+2x)^2 /(x*x)
solve for x
750*x^2 = 0.2^2 + 2*0.2x + 4x^2
(750-4)*x^2 - 0.4x - 0.04 = 0
x = -0.007
so, substitute:
[H2] = 0 - x = 0.007
[Br2] = 0 - x = 0.007
[HBr] = 0.2 + 2x = 0.2-2*0.00705= 0.186
Verify:
Kc = [HBr]^2 / [H2][Br2]
Kc = (0.186^2)/(0.00705*0.00705) = 706
Nearest answer is 0.19
Kc =(0.19^2)/(0.007*0.007) = 736 which is pretty near to 750 (decimals)
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