Equlibrium Constant Kc And Calculating Eq Concentrations

Equlibrium constant Kc and calculating eq concentrations....?

Look at the equation and you see that the stoichiometry is 1:1, ie 1 mole reactants yield 1 mole of products. So, since you know that 0.21 moles of product is produced at equilibrium, that means that 0.21 moles of CH3CH2COOH and 0.21 moles of CH3OH have been used. So, to get the remaining reactants, just subtract 0.21 moles from the starting amounts. Thus, 0.52 - 0.21 = 0.31. And 0.37 - 0.21 = 0.16. To get the amount of H2O at equilibrium, just add 0.21 to the starting amount of 1.2 to arrive at 1.41. That's how those answers were arrived at.

Since the Kc is given in units of concentration, you must use the VOLUME of the reaction and not the actual number of moles. So, that's why each value at equilibrium is divided by V, the volume.

Calculate the equilibrium concentrations of PCl5, PCl3, and Cl2.?

At a certain temperature, the reaction PCl5(g) PCl3(g) + Cl2(g) has an equilibrium constant Kc = 5.8 x 10-2. If only PCl5 is present initially, at a concentration of 0.160 M, calculate the equilibrium concentrations of PCl5, PCl3, and Cl2.

_PCl5
_PCl3
_Cl2

please help im confuse when i attempted to solve it:(

Calculate the concentrations of these two species at equilibrium HeLP! PLEASE?

Br2(g) --> 2Br(g)
Kc = 1.1*10^-3
Kc = [Br]^2 / [Br2]

Q = [0.025]^2 / [0.039]
Q = 0.01608
Q is too big, because its product [Br] is too big,
the reaction will shift to the left

==================================
in:

Br2(g) <= <= <= <= 2Br(g)
0.039 M + x <= <= 0.025 - 2X

Kc = [Br]^2 / [Br2]

1.1 X10^-3 = [0.025 - 2X]^2 / [0.039 M + x]

1.1*10^-3 [0.039 M + x] = [0.025 - 2X]^2

(4.29 X 10^-5) & (1.1 X10^-3x) = 6.25 X 10^-4 - 0.10 X + 4X2

4X2 & [ (6.25 X 10^-4) - (4.29 X 10^-5) ] & [ - 0.10 X - (1.1 X10^-3x) ] = 0

4X2 & [ 0.0005821 ] & [ - 0.1011 X] = 0

using the quadratic at http://www.math.com/students/calculators...
X = 0.00887

I am close to your answer....
quadratic had to be used beacuse X is significantly large

Br2(g) <= <= <= <= 2Br(g)
0.039 M + 0.00887 <= <= 0.025 - 2(0.00887)

[0.0479M] <=> 0.0073 M

in essence , we have the same answer
==========================
as an edit:

although the molarities were different,
this problem was worked by the same method two months ago, & got the right answer http://answers.yahoo.com/question/index;...

that person worked as if it shifted to the right....
but corrected that with a x = a minus value:
(x = -0.00178)

I hink that we might jave the right answer

How to calculate the equilibrium constant for the following reaction??? help please.......?

Firstly calculate the initial concentration of NOCl (C=n/V) gives 0.14M. Now you need the equlibrium concentrations of all substances. So you know that from initial to equilibrium the change in NOCl concentration is (0.14-0.0582)=-0.0818M. As the mole ratio (from equation) of NOCl to NO is 1:1, the change in concentration is the same, 0.0818M of NO is formed. This is concentration at eq as initally no NO was present. For Cl2, mole ratio is 2:1, so change in concn is 0.0409M, which is Cl2 concn at equilibrium.

Now the formula for equilibrium constant is k=equlibrium concentration of NO(squared) multiplied by eq concn Cl2, all divided by eq concn of NOCl squared. K=(0.0818)squared times (0.0409), all divided by (0.0582) squared, which gives you k=0.081

Calculate the value of the equilibrium constant, Kc.?

A mixture initially contains A, B, and C in the following concentrations: [A] = 0.300M , [B] = 1.40M , and [C] = 0.550M . The following reaction occurs and equilibrium is established:


A+2B⇌C

At equilibrium, [A] = 0.200M and [C] = 0.650M .

Equilibrium constant concentration problem? help please....?

Br2(g) <==> 2Br (g)
Initial concentration: 6.3E-2 M and 1.2E-2M respectively
Final concentration: 6.3E2 -x and 1.2E-2 + 2x

Hence, (1.2E-2+2x)^2/(6.3E-2-x) = 1.1E-3

x = -0.00178

[Br] = 1.2E-2 + 2x = 8.4E-3

[Br2] = 6.3E-2 - x = 6.5E-2

If you are using a graphing calculator, set guess = 0.001

Find the equilibrium constant (Kc) For the reaction.?

CH4(g) + 2H2S(g) <-> CS2 (g) + 4 H2 (g)
I .5M .75M 0M 0M
C -.11M -.22M +.11M +.44M
E .39M .53M .11M .44M


you knew the initial molarity for CH4, H2S, and the final molarity of H2

then use the final [H2] to find the changes in [] using stoichiometry--the coefficient of the balanced equation

now you have the equilibrium [], use the Kc equation to find Kc

Kc={(.44)^4*(.11)}/{(.39)*(.53)^2}
=.038

ans. e) .038

hth

Calculate the concentration of each species at equilibrium?

H₂(g) + CO₂(g) ⇄ H₂O(g) + CO(g)

So equilibrium concentrations satisfy the equation
Kc = [H₂O] [CO] / ( [H₂]∙[CO₂] )

Initial reactant concentrations are
[H₂] = [CO₂] = 0.83mol / 5.1L = 0.1627 M ≡ c₀

Setup ICE table
.......... [H₂].......... [CO₂].......... [H₂O].......... [CO]
I........... c₀............. c₀................ 0................ 0
C......... -x.............. -x............... +x.............. +x
E........ c₀ -x...........c₀-x.............. x................ x

So
Kc = [H₂O] [CO] / ( [H₂]∙[CO₂] ) = x²/(c₀ - x)²
<=>
√Kc = x/(c₀ - x)
<=>
x = c₀ /(1 + 1/√Kc)
= 0.1627 M / (1 + 1/√4.2)
= 0.1093 M

The equilibrium concentrations are
[H₂] = c₀ - x = 0.1627 M - 0.1093M = 0.0534 M
[CO₂] = [H₂] = 0.0534 M
[H₂O] = x = 0.1093 M
[CO] = [H₂O] = 0.1093M