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By the end of this section, you will be able to:
  • Derive reaction quotients from chemical equations representing homogeneous and heterogeneous reactions
  • Calculate values of reaction quotients and equilibrium constants, using concentrations and pressures
  • Relate the magnitude of an equilibrium constant to properties of the chemical system

Now that we have a symbol (⇌) to designate reversible reactions, we will need a way to express mathematically how the amounts of reactants and products affect the equilibrium of the system. A general equation for a reversible reaction may be written as follows:

m A + n B + x C + y D

We can write the reaction quotient ( Q )    for this equation. When evaluated using concentrations, it is called Q c . We use brackets to indicate molar concentrations of reactants and products.

Q c = [ C ] x [ D ] y [ A ] m [ B ] n

The reaction quotient is equal to the molar concentrations of the products of the chemical equation (multiplied together) over the reactants (also multiplied together), with each concentration raised to the power of the coefficient of that substance in the balanced chemical equation. For example, the reaction quotient for the reversible reaction 2 NO 2 ( g ) N 2 O 4 ( g ) is given by this expression:

Q c = [ N 2 O 4 ] [ NO 2 ] 2

Writing reaction quotient expressions

Write the expression for the reaction quotient for each of the following reactions:

(a) 3 O 2 ( g ) 2 O 3 ( g )

(b) N 2 ( g ) + 3 H 2 ( g ) 2 NH 3 ( g )

(c) 4 NH 3 ( g ) + 7 O 2 ( g ) 4 NO 2 ( g ) + 6 H 2 O ( g )

Solution

(a) Q c = [ O 3 ] 2 [ O 2 ] 3

(b) Q c = [ NH 3 ] 2 [ N 2 ] [ H 2 ] 3

(c) Q c = [ NO 2 ] 4 [ H 2 O ] 6 [ NH 3 ] 4 [ O 2 ] 7

Check your learning

Write the expression for the reaction quotient for each of the following reactions:

(a) 2 SO 2 ( g ) + O 2 ( g ) 2 SO 3 ( g )

(b) C 4 H 8 ( g ) 2 C 2 H 4 ( g )

(c) 2 C 4 H 10 ( g ) + 13 O 2 ( g ) 8 CO 2 ( g ) + 10 H 2 O ( g )

Answer:

(a) Q c = [ SO 3 ] 2 [ SO 2 ] 2 [ O 2 ] ; (b) Q c = [ C 2 H 4 ] 2 [ C 4 H 8 ] ; (c) Q c = [ CO 2 ] 8 [ H 2 O ] 10 [ C 4 H 10 ] 2 [ O 2 ] 13

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The numeric value of Q c for a given reaction varies; it depends on the concentrations of products and reactants present at the time when Q c is determined. When pure reactants are mixed, Q c is initially zero because there are no products present at that point. As the reaction proceeds, the value of Q c increases as the concentrations of the products increase and the concentrations of the reactants simultaneously decrease ( [link] ). When the reaction reaches equilibrium, the value of the reaction quotient no longer changes because the concentrations no longer change.

Three graphs are shown and labeled, “a,” “b,” and “c.” All three graphs have a vertical dotted line running through the middle labeled, “Equilibrium is reached.” The y-axis on graph a is labeled, “Concentration,” and the x-axis is labeled, “Time.” Three curves are plotted on graph a. The first is labeled, “[ S O subscript 2 ];” this line starts high on the y-axis, ends midway down the y-axis, has a steep initial slope and a more gradual slope as it approaches the far right on the x-axis. The second curve on this graph is labeled, “[ O subscript 2 ];” this line mimics the first except that it starts and ends about fifty percent lower on the y-axis. The third curve is the inverse of the first in shape and is labeled, “[ S O subscript 3 ].” The y-axis on graph b is labeled, “Concentration,” and the x-axis is labeled, “Time.” Three curves are plotted on graph b. The first is labeled, “[ S O subscript 2 ];” this line starts low on the y-axis, ends midway up the y-axis, has a steep initial slope and a more gradual slope as it approaches the far right on the x-axis. The second curve on this graph is labeled, “[ O subscript 2 ];” this line mimics the first except that it ends about fifty percent lower on the y-axis. The third curve is the inverse of the first in shape and is labeled, “[ S O subscript 3 ].” The y-axis on graph c is labeled, “Reaction Quotient,” and the x-axis is labeled, “Time.” A single curve is plotted on graph c. This curve begins at the bottom of the y-axis and rises steeply up near the top of the y-axis, then levels off into a horizontal line. The top point of this line is labeled, “k.”
(a) The change in the concentrations of reactants and products is depicted as the 2 SO 2 ( g ) + O 2 ( g ) 2 SO 3 ( g ) reaction approaches equilibrium. (b) The change in concentrations of reactants and products is depicted as the reaction 2 SO 3 ( g ) 2 SO 2 ( g ) + O 2 ( g ) approaches equilibrium. (c) The graph shows the change in the value of the reaction quotient as the reaction approaches equilibrium.

When a mixture of reactants and products of a reaction reaches equilibrium at a given temperature, its reaction quotient always has the same value. This value is called the equilibrium constant ( K )    of the reaction at that temperature. As for the reaction quotient, when evaluated in terms of concentrations, it is noted as K c .

Questions & Answers

how did you get 1640
Noor Reply
If auger is pair are the roots of equation x2+5x-3=0
Peter Reply
Wayne and Dennis like to ride the bike path from Riverside Park to the beach. Dennis’s speed is seven miles per hour faster than Wayne’s speed, so it takes Wayne 2 hours to ride to the beach while it takes Dennis 1.5 hours for the ride. Find the speed of both bikers.
MATTHEW Reply
420
Sharon
from theory: distance [miles] = speed [mph] × time [hours] info #1 speed_Dennis × 1.5 = speed_Wayne × 2 => speed_Wayne = 0.75 × speed_Dennis (i) info #2 speed_Dennis = speed_Wayne + 7 [mph] (ii) use (i) in (ii) => [...] speed_Dennis = 28 mph speed_Wayne = 21 mph
George
Let W be Wayne's speed in miles per hour and D be Dennis's speed in miles per hour. We know that W + 7 = D and W * 2 = D * 1.5. Substituting the first equation into the second: W * 2 = (W + 7) * 1.5 W * 2 = W * 1.5 + 7 * 1.5 0.5 * W = 7 * 1.5 W = 7 * 3 or 21 W is 21 D = W + 7 D = 21 + 7 D = 28
Salma
Devon is 32 32​​ years older than his son, Milan. The sum of both their ages is 54 54​. Using the variables d d​ and m m​ to represent the ages of Devon and Milan, respectively, write a system of equations to describe this situation. Enter the equations below, separated by a comma.
Aaron Reply
find product (-6m+6) ( 3m²+4m-3)
SIMRAN Reply
-42m²+60m-18
Salma
what is the solution
bill
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bill
-24m+3+3mÁ^2
Susan
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Amira
I only got 42 the rest i don't know how to solve it. Please i need help from anyone to help me improve my solving mathematics please
Amira
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Aphelele
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Bajemah
-6m(3mA²+4m-3)+6(3mA²+4m-3) =-18m²A²-24m²+18m+18mA²+24m-18 Rearrange like items -18m²A²-24m²+42m+18A²-18
Salma
complete the table of valuesfor each given equatio then graph. 1.x+2y=3
Jovelyn Reply
x=3-2y
Salma
y=x+3/2
Salma
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Enock
given that (7x-5):(2+4x)=8:7find the value of x
Nandala
3x-12y=18
Kelvin
please why isn't that the 0is in ten thousand place
Grace Reply
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Grace
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A meditation garden is in the shape of a right triangle, with one leg 7 feet. The length of the hypotenuse is one more than the length of one of the other legs. Find the lengths of the hypotenuse and the other leg
Marry Reply
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state in which quadrant or on which axis each of the following angles given measure. in standard position would lie 89°
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Method
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Enock
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Roger
The Jones family took a 15 mile canoe ride down the Indian River in three hours. After lunch, the return trip back up the river took five hours. Find the rate, in mph, of the canoe in still water and the rate of the current.
cameron Reply
Shakir works at a computer store. His weekly pay will be either a fixed amount, $925, or $500 plus 12% of his total sales. How much should his total sales be for his variable pay option to exceed the fixed amount of $925.
mahnoor Reply
I'm guessing, but it's somewhere around $4335.00 I think
Lewis
12% of sales will need to exceed 925 - 500, or 425 to exceed fixed amount option. What amount of sales does that equal? 425 ÷ (12÷100) = 3541.67. So the answer is sales greater than 3541.67. Check: Sales = 3542 Commission 12%=425.04 Pay = 500 + 425.04 = 925.04. 925.04 > 925.00
Munster
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When traveling to Great Britain, Bethany exchanged $602 US dollars into £515 British pounds. How many pounds did she receive for each US dollar?
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Jazmine trained for 3 hours on Saturday. She ran 8 miles and then biked 24 miles. Her biking speed is 4 mph faster than her running speed. What is her running speed?
Zack Reply
d=r×t the equation would be 8/r+24/r+4=3 worked out
Sheirtina
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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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