#### Dentista08

##### looking for a practice!
10+ Year Member
5+ Year Member
I just dont get where did 2.854 came from? How do we know the concentrations of the products? Please see attachment
The answer is A not b

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#### chessxwizard

##### Full Member
10+ Year Member
7+ Year Member
So in this question, they give you the moles of starting reactant, volume of the container, and the percent of products that are formed. First of all, you need to find the molarity of the reactants:

14.250 moles PCl5 / 3 L container = 4.75M PCl5
And you know that 40% of the reactants are formed into products:

.40 x 4.75 M = 1.9 M of each PCl3 and Cl2 are formed in the reaction.

Then the reaction proceeds in the way written... create an ICE table and plug in the values, keeping everything written in moles/L:

PCl5 -> PCl3 + Cl2
initial 4.75 0 0
change -x x x
equilib. 4.75-x x x

But we know that x = 1.9M, since that's how much is formed at the end of the reaction! So for the Kc value, we just plug in x into the equilibrium values:

PCl5 @ Equilibrium: 4.75-1.9 = 2.85M
PCl3 @ Equilibrium: 1.9M
Cl2 @ Equilibrium: 1.9M

Therefore, Kc = [Cl2][PCl3]/[PCl5]
= [1.9^2]/[2.85], choice A.

#### Dentista08

##### looking for a practice!
10+ Year Member
5+ Year Member
So in this question, they give you the moles of starting reactant, volume of the container, and the percent of products that are formed. First of all, you need to find the molarity of the reactants:

14.250 moles PCl5 / 3 L container = 4.75M PCl5
And you know that 40% of the reactants are formed into products:

.40 x 4.75 M = 1.9 M of each PCl3 and Cl2 are formed in the reaction.

Then the reaction proceeds in the way written... create an ICE table and plug in the values, keeping everything written in moles/L:

PCl5 -> PCl3 + Cl2
initial 4.75 0 0
change -x x x
equilib. 4.75-x x x

But we know that x = 1.9M, since that's how much is formed at the end of the reaction! So for the Kc value, we just plug in x into the equilibrium values:

PCl5 @ Equilibrium: 4.75-1.9 = 2.85M
PCl3 @ Equilibrium: 1.9M
Cl2 @ Equilibrium: 1.9M

Therefore, Kc = [Cl2][PCl3]/[PCl5]
= [1.9^2]/[2.85], choice A.
totally get it now!!! thank you

#### sciencegod

##### Super Member
10+ Year Member
So in this question, they give you the moles of starting reactant, volume of the container, and the percent of products that are formed. First of all, you need to find the molarity of the reactants:

14.250 moles PCl5 / 3 L container = 4.75M PCl5
And you know that 40% of the reactants are formed into products:

.40 x 4.75 M = 1.9 M of each PCl3 and Cl2 are formed in the reaction.

Then the reaction proceeds in the way written... create an ICE table and plug in the values, keeping everything written in moles/L:

PCl5 -> PCl3 + Cl2
initial 4.75 0 0
change -x x x
equilib. 4.75-x x x

But we know that x = 1.9M, since that's how much is formed at the end of the reaction! So for the Kc value, we just plug in x into the equilibrium values:

PCl5 @ Equilibrium: 4.75-1.9 = 2.85M
PCl3 @ Equilibrium: 1.9M
Cl2 @ Equilibrium: 1.9M

Therefore, Kc = [Cl2][PCl3]/[PCl5]
= [1.9^2]/[2.85], choice A.

i dont get it. if 40% gets det decomposed. doesnt that mean that 40% which is 1.9 M should be split into the two products and each one should only be .95?

10+ Year Member
i dont get it. if 40% gets det decomposed. doesnt that mean that 40% which is 1.9 M should be split into the two products and each one should only be .95?

Each mole of PCl5 splits into a mole of PCl3 and Cl2 each. This is basic stoichiometry; think back to the Initial Change Equilibrium (ICE) chart in gen chem.

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