You can't work this problem without knowing the volume. If we assume a volume of 1L, then (H2) initial is 0.1 M and (I2) initial is 0.2 M.
.......H2 + I2 ==> 2HI
I.....0.1M..0.2M....0
C.....-x.....-x.....2x
E...0.1-x...0.2-x...2x
Substitute the E line into the Keq e4xpression and solve for x, then evaluate 0.1-x and 0.2-x.
Calculate the equilibrium amounts of each susbtance in the reaction below if an initial amount of 0.1 moles of H2 are brought together with an initial amount of 0.2 moles of I2 and then equilibrium is established at 900K. Kc at this temperature=70.
H2 + I2 <-> 2HI
3 answers
I'm so sorry I'm still confused
Sabrina, This is probably way more than you need, but you seem very anxious about this concept when it is quite simple in theory. Please don’t stress out over something like this. It’s really bad for you social life… Anyways, Here’s my take on the problem with a little background to support the idea. I hope it helps. Let me know. Doc
Background:
Think of a generalized reaction => Reactants (R) => Products (P) over time. Before the reaction starts, ask yourself, ‘how much reactant and product do you have at time = 0 sec? (100% Reactants & 0% Products) right?
Now, think of a chemical reaction to be like a ‘horse race’ … Before the race starts (or, reaction starts) the horses (reactants) are in the starting gate and no products are forming. Then (mentally) open the gates and let the horses (reactants) run. The concentrations of the reactants will decrease and the concentrations of the products will increase until reaching a point of constant concentration. (Since I can't illustrate this graphically in this format, try to plot the two trends to get a graphic image in your mind’s eye. One graph will be the increasing concentration of the products from 0% to constant concentration. The other graph will show a decrease in concentration of reactants over time until it too reaches a constant concentration. From the start of the race at time = 0, (start of reaction) to the constant concentration point, this is the Kinetic Region of the reaction. Concentrations are constantly changing. At the point when the concentrations become constant, the process enters the ‘Equilibrium Region’ and the amounts of product & reactants remain constant over time because there is a forward reaction and a reverse reaction taking place at the same rate. This is called the ‘Law of Mass Action’. When working chemistry problems in the Kinetic Region, one uses ‘Rate Laws’ to define reaction behavior. When working chemistry problems in the Equilibrium Region, one uses the concept of Equilibrium Constants (derived from Law of Mass Action) that define the ‘Extent of Rxn’. The problem set-up is typically constructed around an ‘I.C.E. Table’ which defines initial concentrations (I) of reactants (horses/reactants in the starting gate), changes in concentration (∆C) of reactants as they proceed to the equilibrium point (Kinetic Region) and Equilibrium Concentrations in the Equilibrium Region (E).
Now, for your problem in equilibrium …
1st) Build an I.C.E. Table around the reaction of interest. Here’s the ICE table for your reaction…
------I₂(g) + H₂(g) <=> --- 2HI(g)
(I) 0.20mole 0.10mole ---- 0.00 mole (This is the starting gate. No products have formed)
(C)-- -x ------ -x ---------- +2x (This is the Kinetic Region. Changes are proportional to coefficients of the balanced equation)
(E) (0.20-x)--(0.10-x)-------- 2x (This is the Equilibrium Region)
2nd) set up the Equilibrium Expression in terms of the equilibrium constant for the problem (Keq).
The Keq-expression must be in ‘Molar’ concentrations. Your problem only specifies moles. They are not the same thing. I’ll assume the reaction is in a 1.00 Liter vessel. This would then give the concentrations in moles/Liter = Molar concentrations.
Rewriting the ICE table in molar terms …
----I₂(g) + H₂(g)-- <=> --- 2HI(g)
(I)0.20M ---0.10M ---------- 0.00M
(C) -x ----- -x ------------- +2x
(E)(0.20-x) (0.10-x) --------- 2x
3rd) Write the Keq = Kc expression for the reaction. Keq is the product of the molar concentrations of the ‘Products’ divided by the product of the molar concentrations of the reactants each raised to their respective coefficient in the balanced chemical equation of interest. That is, … for ...
I₂(g) + H₂(g) <=> 2HI(g)
Kc = [HI]² /[I₂][H₂]
Substitute the equilibrium row (E) Molar Values from the ICE table into the Kc Expression …
Kc = (2x)² /(0.20-x)(0.10-x) = 70
4th) Solve for ‘x’. Your problem will need to use the quadratic equation to obtain x. Once you have the ‘x’ value, apply it to the equilibrium quantities in the ICE table and your problem is solved.
There's my take on this problem...
Kc = (2x)² /(0.20-x)(0.10-x) = 70
=> (+279)x² + (+0.30)x + (- 0.02) = 0
Using the Quadratic Equation, X = [-b ± √b² - 4ac/2a]
Where a = +279, b = +0.30, c = -0.20
X = [-(+0.30) ± √(0.30)² - 4(279)(-0.02)]/[2(279)] = 0.008M
Therefore, Equilibrium concentrations are ...
[I₂(g)] = (0.20 – 0.008)M = 0.192M
[H₂(g)] = (0.10 – 0.008)M = 0.092M
[HI(g)] = 2(0.008)M = 0.016M
Background:
Think of a generalized reaction => Reactants (R) => Products (P) over time. Before the reaction starts, ask yourself, ‘how much reactant and product do you have at time = 0 sec? (100% Reactants & 0% Products) right?
Now, think of a chemical reaction to be like a ‘horse race’ … Before the race starts (or, reaction starts) the horses (reactants) are in the starting gate and no products are forming. Then (mentally) open the gates and let the horses (reactants) run. The concentrations of the reactants will decrease and the concentrations of the products will increase until reaching a point of constant concentration. (Since I can't illustrate this graphically in this format, try to plot the two trends to get a graphic image in your mind’s eye. One graph will be the increasing concentration of the products from 0% to constant concentration. The other graph will show a decrease in concentration of reactants over time until it too reaches a constant concentration. From the start of the race at time = 0, (start of reaction) to the constant concentration point, this is the Kinetic Region of the reaction. Concentrations are constantly changing. At the point when the concentrations become constant, the process enters the ‘Equilibrium Region’ and the amounts of product & reactants remain constant over time because there is a forward reaction and a reverse reaction taking place at the same rate. This is called the ‘Law of Mass Action’. When working chemistry problems in the Kinetic Region, one uses ‘Rate Laws’ to define reaction behavior. When working chemistry problems in the Equilibrium Region, one uses the concept of Equilibrium Constants (derived from Law of Mass Action) that define the ‘Extent of Rxn’. The problem set-up is typically constructed around an ‘I.C.E. Table’ which defines initial concentrations (I) of reactants (horses/reactants in the starting gate), changes in concentration (∆C) of reactants as they proceed to the equilibrium point (Kinetic Region) and Equilibrium Concentrations in the Equilibrium Region (E).
Now, for your problem in equilibrium …
1st) Build an I.C.E. Table around the reaction of interest. Here’s the ICE table for your reaction…
------I₂(g) + H₂(g) <=> --- 2HI(g)
(I) 0.20mole 0.10mole ---- 0.00 mole (This is the starting gate. No products have formed)
(C)-- -x ------ -x ---------- +2x (This is the Kinetic Region. Changes are proportional to coefficients of the balanced equation)
(E) (0.20-x)--(0.10-x)-------- 2x (This is the Equilibrium Region)
2nd) set up the Equilibrium Expression in terms of the equilibrium constant for the problem (Keq).
The Keq-expression must be in ‘Molar’ concentrations. Your problem only specifies moles. They are not the same thing. I’ll assume the reaction is in a 1.00 Liter vessel. This would then give the concentrations in moles/Liter = Molar concentrations.
Rewriting the ICE table in molar terms …
----I₂(g) + H₂(g)-- <=> --- 2HI(g)
(I)0.20M ---0.10M ---------- 0.00M
(C) -x ----- -x ------------- +2x
(E)(0.20-x) (0.10-x) --------- 2x
3rd) Write the Keq = Kc expression for the reaction. Keq is the product of the molar concentrations of the ‘Products’ divided by the product of the molar concentrations of the reactants each raised to their respective coefficient in the balanced chemical equation of interest. That is, … for ...
I₂(g) + H₂(g) <=> 2HI(g)
Kc = [HI]² /[I₂][H₂]
Substitute the equilibrium row (E) Molar Values from the ICE table into the Kc Expression …
Kc = (2x)² /(0.20-x)(0.10-x) = 70
4th) Solve for ‘x’. Your problem will need to use the quadratic equation to obtain x. Once you have the ‘x’ value, apply it to the equilibrium quantities in the ICE table and your problem is solved.
There's my take on this problem...
Kc = (2x)² /(0.20-x)(0.10-x) = 70
=> (+279)x² + (+0.30)x + (- 0.02) = 0
Using the Quadratic Equation, X = [-b ± √b² - 4ac/2a]
Where a = +279, b = +0.30, c = -0.20
X = [-(+0.30) ± √(0.30)² - 4(279)(-0.02)]/[2(279)] = 0.008M
Therefore, Equilibrium concentrations are ...
[I₂(g)] = (0.20 – 0.008)M = 0.192M
[H₂(g)] = (0.10 – 0.008)M = 0.092M
[HI(g)] = 2(0.008)M = 0.016M
1 answer