# chemistry aqa a level equilibrium question

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sierranonis

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could someone explain 4b to me please?

Q4.

When substances P and Q react together to form substance R an equilibrium is established

according to the equation

P(g) + Q(g) ⇌ 2R(g)

The equilibrium constant expression is K c = *

1.0 mol of P and 1.0 mol of Q were mixed in a container with volume 1.0 dm 3

At equilibrium, x mol of P had reacted.

(a)*****The amount, in moles, of each of P and Q at equilibrium is (1− x).

Deduce in terms of x the amount, in moles, of R in the equilibrium mixture.

___________________________________________________________________

(1)

(b)*****At 298 K the value of the equilibrium constant K c = 3.6

Calculate a value for the equilibrium concentration, in mol dm −3 , of R.

Q4.

When substances P and Q react together to form substance R an equilibrium is established

according to the equation

P(g) + Q(g) ⇌ 2R(g)

The equilibrium constant expression is K c = *

1.0 mol of P and 1.0 mol of Q were mixed in a container with volume 1.0 dm 3

At equilibrium, x mol of P had reacted.

(a)*****The amount, in moles, of each of P and Q at equilibrium is (1− x).

Deduce in terms of x the amount, in moles, of R in the equilibrium mixture.

___________________________________________________________________

(1)

(b)*****At 298 K the value of the equilibrium constant K c = 3.6

Calculate a value for the equilibrium concentration, in mol dm −3 , of R.

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Isame

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okay so kc = 3.6

R is equal to 2x from part a)

3.6 = [2x] / [1-x]^2

(1-x)(1-x) = 1-2x x^2

3.6 = 1-2x x^2

- (1-2x x^2) from both sides

-x^2 2x 2.6 = 0

quadratic formula = -(2) - √(2)^2 - 4(-1)(2.6) / 2(-1)

x= 2.9, but its 2x so 5.8

If im completely wrong i'm sorry, I dont take maths

R is equal to 2x from part a)

3.6 = [2x] / [1-x]^2

(1-x)(1-x) = 1-2x x^2

3.6 = 1-2x x^2

- (1-2x x^2) from both sides

-x^2 2x 2.6 = 0

quadratic formula = -(2) - √(2)^2 - 4(-1)(2.6) / 2(-1)

x= 2.9, but its 2x so 5.8

If im completely wrong i'm sorry, I dont take maths

Last edited by Isame; 10 months ago

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BDavies1

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(Original post by

could someone explain 4b to me please?

Q4.

When substances P and Q react together to form substance R an equilibrium is established

according to the equation

P(g) + Q(g) ⇌ 2R(g)

The equilibrium constant expression is K c = *

1.0 mol of P and 1.0 mol of Q were mixed in a container with volume 1.0 dm 3

At equilibrium, x mol of P had reacted.

(a)*****The amount, in moles, of each of P and Q at equilibrium is (1− x).

Deduce in terms of x the amount, in moles, of R in the equilibrium mixture.

___________________________________________________________________

(1)

(b)*****At 298 K the value of the equilibrium constant K c = 3.6

Calculate a value for the equilibrium concentration, in mol dm −3 , of R.

**sierranonis**)could someone explain 4b to me please?

Q4.

When substances P and Q react together to form substance R an equilibrium is established

according to the equation

P(g) + Q(g) ⇌ 2R(g)

The equilibrium constant expression is K c = *

1.0 mol of P and 1.0 mol of Q were mixed in a container with volume 1.0 dm 3

At equilibrium, x mol of P had reacted.

(a)*****The amount, in moles, of each of P and Q at equilibrium is (1− x).

Deduce in terms of x the amount, in moles, of R in the equilibrium mixture.

___________________________________________________________________

(1)

(b)*****At 298 K the value of the equilibrium constant K c = 3.6

Calculate a value for the equilibrium concentration, in mol dm −3 , of R.

b. Kc= {(2x)^2}/V/ {(1-x)^2}/V^2 ... where V is the volume...because same no. of moles on the left and the right, V's will cancel out so:

Kc=(2x)^2/(1-x)^2

either: multiply out, you will get a quadratic equation ... it won't factorise easily so you will have to use the quadratic formula

or: since it is a perfect square, take square roots of both sides of equation

sqrt(3.6) = 2x/(1-x)

so 1.897 = 2x/(1-x)

rearrange and solve for x, x= 0.487 moles at eqm

This is off an AQA paper. They don't want you to use the quadratic formula hence they make it possible to use square root method

Note this is quite an artificial situation- it is usually hard to find concentrations given Kc value (much easier the other way around) because unless you have equal numbers of moles on both sides of the equation the algebra gets very difficult- this is a chemistry exam and they want to examine chemistry not algebra.

Note, that if you had not started with the same moles (1) of P and Q that this square root method would not work.

(see "Davies A-level chemistry" you tube search for Kc

They have asked q's of this square root method twice I think (one of them may have been on a specimen paper)

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sierranonis

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#4

(Original post by

okay so kc = 3.6

R is equal to 2x from part a)

3.6 = [2x] / [1-x]^2

(1-x)(1-x) = 1-2x x^2

3.6 = 1-2x x^2

- (1-2x x^2) from both sides

-x^2 2x 2.6 = 0

quadratic formula = -(2) - √(2)^2 - 4(-1)(2.6) / 2(-1)

x= 2.9, but its 2x so 5.8

If im completely wrong i'm sorry, I dont take maths

**Isame**)okay so kc = 3.6

R is equal to 2x from part a)

3.6 = [2x] / [1-x]^2

(1-x)(1-x) = 1-2x x^2

3.6 = 1-2x x^2

- (1-2x x^2) from both sides

-x^2 2x 2.6 = 0

quadratic formula = -(2) - √(2)^2 - 4(-1)(2.6) / 2(-1)

x= 2.9, but its 2x so 5.8

If im completely wrong i'm sorry, I dont take maths

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sierranonis

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#5

**Isame**)

okay so kc = 3.6

R is equal to 2x from part a)

3.6 = [2x] / [1-x]^2

(1-x)(1-x) = 1-2x x^2

3.6 = 1-2x x^2

- (1-2x x^2) from both sides

-x^2 2x 2.6 = 0

quadratic formula = -(2) - √(2)^2 - 4(-1)(2.6) / 2(-1)

x= 2.9, but its 2x so 5.8

If im completely wrong i'm sorry, I dont take maths

(Original post by

a. there are 2x moles of R at eqm

b. Kc= {(2x)^2}/V/ {(1-x)^2}/V^2 ... where V is the volume...because same no. of moles on the left and the right, V's will cancel out so:

Kc=(2x)^2/(1-x)^2

either: multiply out, you will get a quadratic equation ... it won't factorise easily so you will have to use the quadratic formula

or: since it is a perfect square, take square roots of both sides of equation

sqrt(3.6) = 2x/(1-x)

so 1.897 = 2x/(1-x)

rearrange and solve for x, x= 0.487 moles at eqm

This is off an AQA paper. They don't want you to use the quadratic formula hence they make it possible to use square root method

Note this is quite an artificial situation- it is usually hard to find concentrations given Kc value (much easier the other way around) because unless you have equal numbers of moles on both sides of the equation the algebra gets very difficult- this is a chemistry exam and they want to examine chemistry not algebra.

Note, that if you had not started with the same moles (1) of P and Q that this square root method would not work.

(see "Davies A-level chemistry" you tube search for Kc

They have asked q's of this square root method twice I think (one of them may have been on a specimen paper)

**BDavies1**)a. there are 2x moles of R at eqm

b. Kc= {(2x)^2}/V/ {(1-x)^2}/V^2 ... where V is the volume...because same no. of moles on the left and the right, V's will cancel out so:

Kc=(2x)^2/(1-x)^2

either: multiply out, you will get a quadratic equation ... it won't factorise easily so you will have to use the quadratic formula

or: since it is a perfect square, take square roots of both sides of equation

sqrt(3.6) = 2x/(1-x)

so 1.897 = 2x/(1-x)

rearrange and solve for x, x= 0.487 moles at eqm

This is off an AQA paper. They don't want you to use the quadratic formula hence they make it possible to use square root method

Note this is quite an artificial situation- it is usually hard to find concentrations given Kc value (much easier the other way around) because unless you have equal numbers of moles on both sides of the equation the algebra gets very difficult- this is a chemistry exam and they want to examine chemistry not algebra.

Note, that if you had not started with the same moles (1) of P and Q that this square root method would not work.

(see "Davies A-level chemistry" you tube search for Kc

They have asked q's of this square root method twice I think (one of them may have been on a specimen paper)

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