equilibrium student 2014 2

Chemical Equilibrium

When a system is at equilibrium, the forward and reverse reaction are proceeding at the same rate

The concentrations of all species remains constant over time, but both the forward and reverse reaction never cease

That means this is a dynamic equilibrium, not like a beach ball on a seal’s nose.

Equilibrium is signified with double arrows or the equal sign

Symbol found in “Arrows” section of Insert equations in Microsoft Office.

⇌Equilibrium is signified with double arrows as above Symbol found in “Arrows” section of Insert “Equations” in Microsoft Office.If you cannot find that symbol in your software an acceptable alternative is:

Dynamic Equilibrium

N2O4 2 NO2

Initially forward reaction rapid

As some reacts [N2O4] so rate forward

Initially Reverse reaction slow No products

As NO2 forms Reverse rate

molecules collide more often as [NO2]

Eventually rateforward =ratereverse this is equilibrium

From this point on concentrations will not change

Reversibility N2O4 2 NO2

For given overall system compositionAlways reach same equilibrium concentrationsWhether equilibrium is approached from forward or reverse direction

Mass Action Expression

Relates [reactants] and [products]

Forward reaction A → B Rate = kf[A]

Reverse reaction B → A Rate = kr[B]

Equilibrium A B: kf[A] = kr[B]

Rearranging [A] = kf = constant

[B] kr

6

Mass Action ExpressionUses stoichiometric coefficients as exponent for

each reactant

For reaction: aA + bB cC + dD

Reaction quotientNumerical value of mass action expression

Equals “Q ” at any time, and

Equals “K ” only when reaction is known to be at equilibrium

Calculation Results

Q = __[HI]2_ = same for all = K if all

[H2][I2] experiments at equilibrium

Expt [H2] [I2] [HI] Kc

I .0222 .0222 .156 49.4

II .0350 .0450 .280 49.8

III .0150 .0135 .100 49.4

IV .0442 .0442 .311 49.5

H2(g) + I2(g) 2 HI(g)

Equilibrium established

Equilibrium Law

At 440oC equilibrium law:

Kc = _[HI]2_ = 49.5

[H2][I2]

Equilibrium constant = Kc at given T

Use Kc as normally working with molarity

At equilibrium Q = Kc

Predicting Equilibrium Law

aA + bB cC + dD

Q = [C]c[D]d

[A]a[B]b

Exponents are stoichiometric coefficients of balanced equation.

Law is:

Kc = [C]c[D]d

[A]a[B]b

Predicting Equilibrium Law

Where only concentrations that satisfy this equation are equilibrium concentrations

Numerator

Multiply [products] raised to their stoichiometric coefficients

Denominator

Multiply [reactants] raised to their stoichiometric coefficients

Kc = [products]f

[reactants]d

Check

3H2(g) + N2(g) 2NH3(g)

Kc = 4.26 × 108 at 25 °C

What is equilibrium law?

Check

3H2(g) + N2(g) 2NH3(g)

Kc = 4.26 × 108 at 25 °C

What is equilibrium law?

Example

In the reaction:

N2O4 2 NO2

Q = [NO2]2

[N2O4]

CheckWhich of the following is the correct mass action expression for the reaction:

Cu2+(aq) + 4NH3(aq) [Cu(NH3)42+](aq)?

15

Manipulating Equations

Reverse direction, new Kc is 1/Kc

A + B C + D Kc = [C][D]

[A][B]

C + D A + B K’c = [A][B} = _1_

[C][D] Kc

Reverse Equilibrium

3 H2(g) + N2(g) 2 NH3(g)

Kc = _[NH3]2_ = 4.26 x 108

[H2]3[N2]

2 NH3(g) 3 H2(g) + N2(g)

Kc = [H2]3[N2] = ___1____ = 2.35 x 10-9

[NH3]2 4.26 x 108

Kc Related to Equations II

With aA + bB cC + dD

Kc = [C]c[D]d

[A]a[B]b

If multiply equation, raise Kc to that power

2aA + 2bB 2cC + 2dD

Knew = [C]2c[D]2d

[A]2a[B]2b

Here Knew = Kc2

Kc Related to Equations III

Add equations, multiply Kc values

aA bB K1 = [B]b

[A]a

bB cC K2 = [C]c

[B]b

Add equations: aA cC

Multiply K: Kc = [B]b x [C]c = K1 x K2

[A]a [B]b

Example

3 H2(g) + N2(g) 2 NH3(g)

Kc = _[NH3]2_ = 4.26 x 108

[H2]3[N2]

Multiply by 3

9 H2(g) + 3 N2(g) 6 NH3(g)

Kc = _[NH3]6_ = Kc3 7.73 x 1025

[H2]9[N2]3

Adding Equations

A + B C + D Kc1 = [C][D]

[A][B]

C + E F + G Kc2 = [F][G]

[C][E]

A + B+ E D + F + G

Kcc = [C][D] x [F][G] = [D][F][G]

[A][B] [C][E] [A][B][E]

Adding Equations

For related reactions in series, the overall Kc is obtained:-

Kcc = Kc1 x Kc2

A. 2CO + O2 2CO2 KcA = 3.3 x 1091

B. 2H2 + O2 2H2O KcB = 9.1 x 1080

Find Kc for H2O + CO CO2 + H2

CO + ½O2 CO2

H2O H2 + ½O2 Add:

H2O + CO CO2 + H2

Adding Equations II

Divide A by 2 CO + ½O2 CO2 Take sq. root of K cA

Kc1 = [CO2] = (3.3 x 1091)½ = 5.7 x 104

[CO][O2]½

Divide B by 2 and reverse:H2O H2 + ½O2

Kc2 = [H2][O2]½ = (9.1 x 1080)-½ 3.3 x 10-41 [H2O]

Adding Equations III

Adding equations gives the overall equation and multiplying K values gives the overall Kc:-Kc = [CO2] x [H2][O2]½ = 1.9 x 105

[CO][O2]½ [H2O]

Kc = (5.7 x 1045) x (3.3 x 10-41) = 1.9 x 105

Practice 1

For: N2(g) + 3 H2(g) 2 NH3(g)

Kc = 500. at a particular temperature.

What would be Kc for following?

2 NH3(g) N2(g) + 3 H2(g)

½ N2(g) + 3/2 H2(g) NH3(g)

Practice 2

Equilibrium Constant, Kc

Constant value equal to ratio of product concentrations to reactant concentrations raised to their respective exponents

Kc = [products]f

[reactants]d

Changes with temperature (van’t Hoff Equation)

Depends on solution concentrations

Assumes reactants and products are in solution

Equilibrium Constant, Kp

Based on reactions in which substances are gaseous

Assumes gas quantities are expressed in atmospheres in mass action expression

Use partial pressures for each gas in place of concentrations

Ex. N2 (g) + 3 H2 (g) 2 NH3 (g)

Kp = _P2NH3

PN2.P3

29

How are Kp and Kc Related?Start with ideal gas law

PV = nRT

Rearranging gives

Substituting P/RT for molar concentration into Kc results in pressure-based formula

∆n = moles of gas in product – moles of gas in reactant

Kp = Kc when Δn = 0

Check

Consider the reaction A(g) + 2B(g) 4C(g) If the Kc for the reaction is 0.99 at 25ºC, what would be the Kp?

A. 0.99

B. 2.0

C. 24.

D. 2400

E. None of these

Solution

Consider the reaction A(g) + 2B(g) 4C(g) If the Kc for the reaction is 0.99 at 25ºC, what would be the Kp?

n = nproducts – nreactants = 4 – 3 = 1

Kp = Kc(RT)Δn = 0.99 x (0.08206 x 298)1 = 24.2

Check 2

Consider the reaction: 2NO2(g) N2O4(g)

If Kp = 0.480 for the reaction at 25°C, what is value of Kc at same temperature?

Solution 2

Consider the reaction: 2NO2(g) N2O4(g)

If Kp = 0.480 for the reaction at 25°C, what is value of Kc at same temperature?

n = nproducts – nreactants = 1 – 2 = –1

Kp = Kc(RT)Δng

Kc = _Kp__ = _____0.480___ = 19.6

(RT)Δn (0.08206 x 298)-1

Practice 3

Calculate the value of Kp or Kc for each of the following at 27 °C2 SO2(g) + O2(g) 2 SO3(g) Kc = 8 x 1025

N2(g) + 2 O2(g) 2 NO2(g) Kp = 3 x 10−17

Heterogeneous Equilibrium

Homogeneous reaction/equilibrium

All reactants and products in same phase

Can mix freely

Heterogeneous reaction/equilibrium

Reactants and products in different phases

Can’t mix freely

Solutions are expressed in M

Gases are expressed in M

Governed by Kc

Heterogeneous Equilibrium II

Pure solids and pure liquids changing concentration would be changing density, cannot do that.Therefore solids and liquids are not included in the equilibrium constant expression

For aA(s) + bB(aq) cC(l) + dD(aq)

the equilibrium constant expression is

Kc = [C]c[B]b = [D]d

[A]a[D]d [B]b

Heterogeneous Equilibrium III

2NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g)

Kc = [Na2CO3(s)] [H2Og][CO2(g)] = [H2Og][CO2(g)]

[NaHCO3(s)]2

1 mol NaHCO3 , V = 38.9 mL

M = 1 mol/.0389 L = 25.7 M

2 mol NaHCO3 , V = 77.8 mL

M = 2 mol/.0778 L = 25.7 M

Check

Write equilibrium laws for the following:

Ag+(aq) + Cl–

(aq) AgCl(s)

Kc = ____1___

[Ag+][Cl-]H3PO4(aq) + H2O(ℓ) H3O+

(aq) + H2PO4(aq)

Kc = [H3O+][H2PO4-]

[H3PO4]

CheckGiven the reaction:

3Ca2+(aq) + 2PO43–(aq) Ca3(PO4)2(s)

What is the mass action expression?

39

Large Kc

Kc >> 1

Rich in product

Goes toward complete

2SO2(g) + O2(g) 2SO3(g)

Kc = 7.0 1025 at 25 ° C

Small Kc

Kc << 1

Rich in reactant

Few products

H2(g) + Br2(g) 2HBr(g)

Kc = 1.4 10–21 at 25 °C

K near 1

K 1

Products and reactants

concentratrions close

2NO2(g) N2O4(g)

Kp = 0.480 at 25°C

43

Check

Consider the reaction of 2NO2(g) N2O4(g)

If Kp = 0.480 at 25 °C, does the reaction favor product or reactant?

Equilibrium and Changes

Equilibrium positions Combination of concentrations that allow Q = K

Infinite number of possible equilibrium positions

Le Châtelier’s principleSystem at equilibrium (Q = K) when upset by disturbance (Q ≠ K) will shift to offset stress

Reaction Direction

Can us value of Q to predict reaction direction1. Qc > Kc reaction goes to left; reverse

2. Qc < Kc reaction goes to right; forward

3. Qc = Kc reaction at equilibrium.4. If only reactants Q = 0; forward5. If only reactants Q = ∞; reverse

46

Relationship Between Q and K

Q = K reaction at equilibrium

Q < K reactants go to products

Too many reactants

Must convert some reactant to product to move reaction toward equilibrium: shift to right

Q > K products go to reactantsToo many products

Must convert some product to reactant to move reaction toward equilibrium: shift to left

Le Châtelier’s Principle

1. Concentration

2. Pressure and volume

3. Temperature

4. Catalysts

5. Adding inert gas to system at constant volume

48

1. Effect of Change in ConcentrationCu(H2O)4

2+(aq) + 4Cl–(aq) CuCl42–(aq) + 4H2O

blue yellow

Equilibrium mixture is blue-green

Add excess Cl– (conc HCl)Equilibrium shifts to products

Makes more yellow CuCl42–

Solution becomes green

4)()(

242

42)(

24

]Cl][O)Cu(H[

]OH][CuCl[

aqaq

aqcK

4)()(

242

)(24

42 ]Cl][O)Cu(H[

]CuCl[

]OH[ aqaq

aqcc

KK

49

1. Effect of Change in ConcentrationCu(H2O)4

2+(aq) + 4Cl–(aq) CuCl42–(aq) + 4H2O

blue yellow

Add Ag+ Removes Cl–: Ag+(aq) + Cl–(aq) AgCl(s)

Equilibrium shifts to reactants

Makes more blue Cu(H2O)42+

Solution becomes bluer

Add H2O?

4)()(

242

)(24

]Cl][O)Cu(H[

]CuCl[

aqaq

aqcK

Addition of Product

Add NO2(g) to N2O4(g) 2NO2(g)

System is disturbed so Q now > K; system responds by converting more NO2(g) into N2O4(g) until again Q = K but with larger values for all concentrations.

Check

2SO2(g) + O2(g) → 2SO3(g)

Kc = 2.4 x 10-3 at 700 oC

Which direction will the reaction move if 0.125 moles of O2 is added to an equilibrium mixture ?

A. Towards the products

B. Towards the reactants

C. No change will occur

Concentration Effect Summary

When changing concentrations of reactants or products

Equilibrium shifts to remove reactants or products that have been added

Equilibrium shifts to replace reactants or products that have been removed

Effect of Change in P & V

Gaseous system at constant T and n

3H2 (g) + N2(g) 2NH3 (g) Kp = __P2NH3__

PN2.P3H2

If Vexpect PTo reduce P, look at each side of reaction

Which has less moles of gas

Reactants = 3 + 1 = 4 moles gas

Products = 2 moles gas

Reaction favors products (shifts to right)

Effect of Change in P & V

If Vexpect PTo reduce P, look at each side of reaction

Which has less moles of gas

Reactants = 3 + 1 = 4 moles gas

Products = 2 moles gas

Reaction favors products (shifts to right)

Effect of Change in P & V II

2NaHSO3(s) NaSO3(s) + H2O(g) + SO2(g) Kp = PH2O.PSO2

If you V of reaction,

Reactants: no moles gas = all solids

Products: 2 moles gas

V, causes P

Reaction shifts to left (reactants), as this has fewer moles of gas

Effect of Change in Temperature

Cu(H2O)42+

(aq) + 4Cl–(aq) + Heat CuCl4

2–(aq) + 4H2O

blue yellow

Reaction endothermic

Adding heat shifts equilibrium toward products

Cooling shifts equilibrium toward reactants56

Temperature Change

H2O(s) H2O(ℓ) H° =+6 kJ (at 0 °C)

Energy + H2O(s) H2O(ℓ)

Energy is reactant

Add heat, shift reaction right

3H2(g) + N2(g) 2NH3(g) Hf°= –47.19 kJ

3 H2(g) + N2(g) 2 NH3(g) + energy

Energy is product

Add heat, shift reaction left

Effect of T Change

T shifts reaction in direction that produces endothermic (heat absorbing) change

T shifts reaction in direction that produces exothermic (heat releasing) change

Changes in T change value of mass action expression at equilibrium, so K changed

Effect of T Change II

K depends on T

T of exothermic reaction makes K smallerMore heat (product) forces equilibrium to reactants

T of endothermic reaction makes K largerMore heat (reactant) forces equilibrium to products

Catalysts

Catalysts lower Ea

Change in Ea has

equal effect on kf & kr

Result is no effect

on K

Addition of Inert Gas

Will not affect concentrations or partial pressures of any components

Will not affect reaction

62

Learning Check:Consider:

H3PO4(aq) + 3OH–(aq) 3H2O(ℓ) + PO43–(aq)

What will happen if PO43– is removed?

Q is proportional to [PO43– ]

[PO43– ], Q

Q < K equilibrium shifts to right

]POH[]OH[

]PO[Q

433

34

63

Learning Check:The reaction

H3PO4(aq) + 3OH–(aq) 3H2O(aq) + PO43–(aq)

is exothermic.

What will happen if system is cooled?

Since reaction is exothermic, heat is product Heat is directly proportional to Q T, Q Q < K equilibrium shifts to right

heat

]POH[]OH[

]PO[Q

433

34

Check

The equilibrium between aqueous cobalt ion and the chlorine ion is shown:[Co(H2O)6]2+(aq) + 4Cl–(aq) [Co(Cl)4]2–(aq) + 6H2O(ℓ)

pink blue

It is noted that heating a pink sample causes it to turn violet. The reaction is:

A. endothermic B. exothermic

C. cannot tell from the given information

Calculating Kc from [Equilibrium]

Most direct way.

Measure all reactant & product concentrations.

Kc always be the same as long as T constant

65

Tro: Chemistry: A Molecular Approach, 2/e

Equilibrium Calculations

For solution reactions, must use KC

For gaseous reactions, use either KP or KC

Calculate K from equilibrium [X] or PX

Find one or more equilibrium [X] or PX

from KP or KC

Kc From [X}s at Equilibrium

Use mass action expression to relate [X] values

Can use:Equilibrium [reactant]s & [product]s

From initial and one final [product] find remaining {X}s then find Kc

Kc from Equilibrium [X]

1 N2O4(g) 2NO2(g)

If you place 0.0350 mol N2O4 in 1 L flask at equilibrium, what is KC?

[N2O4]eq = 0.0292 M

[NO2]eq = 0.0116 M

Kc = _[NO2}2 = (0.0292)2 = 4.61 x 10-3

[N2O4] (0.0292)

Practice 4

For the reaction: 2A(aq) + B(aq) 3C(aq) the equilibrium concentrations are: A = 2.0 M, B = 1.0 M and C = 3.0 M. What is the expected value of Kc at this temperature?

A. 14

B. 0.15

C. 1.5

D. 6.75

From [X]i and [Y]f

Set up ICE table

Changes in same ratio as coefficients

[X]eq = [X]initial - [X]change

1. Assign x as coefficient, + for materials on side it is going to or – for opposite

2. Solve for x, if 2nd order, take square root, use quadratic equation or simplify if K very large or very small.

Generic

Use concentration tables set up as follows:

A + B 2C

[A] [B] [C]

I P Q R

C -x -x +2x

E P – x Q – x R + 2x

Kc = _ (R + 2x)2__

(P - x)(Q - x)

Example

2SO2(g) + O2(g) 2SO3(g)

1.000 mol SO2 and 1.000 mol O2 are placed in a 1.000 L flask at 1000 K. At equilibrium 0.925 mol SO3 has formed. Calculate KC for this reaction.

[SO2]i = [O2]i = 1.00 mol = 1.00 M

1.00 L

[SO3]eq = 0.925 mol = 0.925 M

1.00 L

ICE Table

2SO2(g) + O2(g) 2SO3(g)

I 1.00 1.00 0.000

C -0.925 -0.462 +0.925

E 0.075 0.538 0.925

Kc = __[SO3]2__ = _(0.925)2__ 2.8 x 102

[SO2]2[O2] (.075)2(.538)

Practice 5

CH4(g) + H2O(g) CO(g) + 3H2(g) Kc = 5.67

[CH4] = 0.400 M; [H2] = 0.800M; [CO] =0.300M

What is [H2O] at equilibrium?

Practice

H2(g) + I2(g) 2HI(g) at 425 °C

KC = 55.64

If one mole each of H2 and I2 are placed in a 0.500 L flask at 425 °C, what are the equilibrium concentrations of H2, I2 and HI?

Kc = __[HI]2_= 55.64

[H2][I2]

[H2]i = [I2]i = 1.00 mol = 2.00 M

0.500 L

Solution

H2(g) + I2(g) 2HI(g)

I 2.00 2.00 0.00

C -x -x +2x

E 2.00 – x2.00 – x 2x

55.64 = ____(2x)2______ = ____(2x)2___

(2.00 – x)(2.00 – x) (2.00 – x)2

Take square root of both sides

Solution II

55.64 = 7.459 = __(2x)__

(2.00 – x)

Rearrange: (7.459)(2.00 – x) = 2x

14.918 – 7.459x = 2x

14.918 = 9.459x

x = 14.918 = 1.58

9.459

[H2]eq = [I2]eq = 2.00 – 1.58 = 0.42 M

[HI]eq = 2x = 2(1.58) = 3.16 M

Example 2

H2(g) + I2(g) 2HI(g) at 425 °C

KC = 55.64

If one mole each of H2, I2 and HI are placed in a 0.500 L flask at 425 °C, what are the equilibrium concentrations of H2, I2 and HI?

Now have product as well as reactants initially

Kc = __[HI]2_= 55.64

[H2][I2]

H2]i = [I2]i = [HI] 1.00 mol = 2.00 M

0.500 L

Solution

H2(g) + I2(g) 2HI(g)

I 2.00 2.00 2.00

C -x -x +2x

E 2.00 – x2.00 – x 2.00 + 2x

55.64 = __(2.00 + 2x)2__ = (2.00 + 2x)2

(2.00 – x)(2.00 – x) (2.00 – x)2

Take square root of both sides

Solution

55.64 = 7.459 = (2.00 + 2x)_

(2.00 – x)

Rearrange: (7.459)(2.00 – x) = 2.00 + 2x

14.918 – 7.459x = 2.00 + 2x

12.918 = 9.459x

x = 12.918 = 1.37

9.459

[H2]eq = [I2]eq = 2.00 – 1.37 = 0.63 M

[HI]eq = 2.00 + 2(1.37) = 2.00 + 2.74 = 4.74 M

Practice 6

N2(g) + O2(g) → 2NO(g)

Kc = 0.0123 at 3900 oC

If 0.25 moles of N2 and O2 are placed in a 250 mL container, what are the equilibrium concentrations of all species ?

A. 0.0526 M, 0.947 M, 0.105 M

B. 0.947 M, 0.947 M, 0.105 M

C. 0.947 M 0.105 M, 0.0526 M

D. 0.105 M, 0.105 M, 0.947 M

Practice

CH3CO2H(aq) + C2H5OH(aq) CH3CO2C2H5(aq) +H2O(l)

acetic acid ethanol ethyl acetate KC = 0.11

An aqueous solution of ethanol and acetic acid, each with initial concentration of 0.810 M, is heated at 100 °C. What are the concentrations of acetic acid, ethanol and ethyl acetate at equilibrium?

Solution

Kc = __[CH3CO2C2H5]__

[CH3CO2H][C2H5OH]

CH3CO2H C2H5OH CH3CO2C2H5

I 0.810 0.810 0.000

C -x -x +x

E .810 – x .810 – x x

0.11 = ________x________

(.810 – x)(.810 – x)

Solution II

Rearranging:

0.11 x(0.6561 – 1.62x + x2) = x

Expand then write as quadratic equation

ax2 + bx + c = 0

0.07217 – 0.1782x + 0.11x2 - x = 0

0.11x2 – 1.1782x + 0.07217 = 0

x = -b (b2 – 4ac)

2a

85

Solution III

This gives two roots: x = 10.6 and x = 0.064Only x = 0.064 is possible

x = 10.6 is >> 0.810 initial concentrations 0.810 – 10.6 = negative concentration,

which is impossible

)11.0(2)07217.0)(11.0(4)1782.1()1782.1( 2

x

22.0164.11782.1

22.0)032.0()388.1(1782.1

x

Solution IV

[CH3CO2C2H5]eq = x = 0.064 M

[CH3CO2H] = [C2H5OH]

= 0.810 – x = 0.810 – 0.064 = 0.746 M

Simplifications

If get trinomial use successive approximations

If K is very small x is also small can drop x in change part only

Valid if [X]I = 400x > K

More Problems

1.Finding Kc from change in reactant concentrations.

PCl 3(g) + Cl2(g) PCl5(g)

0.200 mol PCl3 and 0.100 mol Cl2 put in 1.00 L flask, at equilibrium found 0.120 mol PCl3. Means change in [PCl3] is 0.200 - 0.120 = 0.080 mol, change must be same for other components. As volume is 1.0 L, molarity has same value.

Example

Using Kc to calculate equilibrium concentrations:A flask contains a mixture of Br2 and Cl2 ,both with an initial concentration of 0.0250 M. Find the equilibrium concentration of all molecules from the following equilibrium information.

Example Continued

[BrCl] [Br2] [Cl2]

I 0 .0250 .0250

C +2x -x -x

E 2x .0250 – x .0250 - x Kc = [Br2][Cl2] = (.0250 - x)2 = .145 [BrCl]2 (2x)2

Continued

Kc = [Br2][Cl2] = (.0250 - x)2 = .145

[BrCl]2 (2x)2

Take square root:(.0250 - x) = .381

2x

.0250 - x = .762 or .0250 = 1.762 x

x = .0250/1.762 = .0142 for Br2 & Cl2

BrCl = 2x = .0284 M

Practice

Find equilibrium concentrations with very small Kc.

N2O + NO2 3NO Kc = 1.4 x 10-10 Start with .200 mol N2O & .400 mol NO2 in

4.00 L

Solution

[N2O] = .200 mol/4.00 L = 0.0500 M

[N2O] = .400 mol/4.00 L = 0.100 M

[N2O] [NO2] [NO]I .0500 .100 0C -x -x +3xE .0500 – x .100 – x 3x

Problem 3 (II)

Kc = [NO]3 = (3x)3 =1.4 x10-10

[N2O][NO2] (.0500 - x)(.100 - x) Simplify reactant concentrations as x very small:

27x3 = 1.4 x 10-10

(.0500)(.100)x3 = 2.6 x 10-14 or x = 3.0 x 10 -5

[NO] = 3x = 9.0 x 10-5