For the reaction: H2+Br2 2HBr, Kc = 7.5x10^2 at a certain temperature. 1.00 mole

Question

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 :)

Explanation / Answer

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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