# Predicting the Direction of a Reaction

Submitted by ChemPRIME Staff on Thu, 12/16/2010 - 14:44

Often you will know the concentrations of reactants and products for a particular reaction and want to know whether the system is at equilibriumA state in which no net change is occurring, that is, in which the concentrations of reactants and products remain constant; chemical equilibrium is characterized by forward and reverse reactions occurring at the same rate.. If it is not, it is useful to predict how those concentrations will change as the reaction approaches equilibrium. A useful tool for making such predictions is the reaction quotient, Q. Q has the same mathematical form as the equilibrium-constant expression, but Q is a ratio of the actual concentrations (not a ratio of equilibrium concentrations).

For example, suppose you are interested in the reaction

2 SO2(g)  +  O2(g)  $\rightleftharpoons$   2 SO3(g)        ${K_{\text{c}}} = \frac{{{{\left[ {{\text{S}}{{\text{O}}_{\text{3}}}} \right]}^2}}}{{{{\left[ {{\text{S}}{{\text{O}}_{\text{2}}}} \right]}^2}\left[ {{{\text{O}}_2}} \right]}} = 245{\text{ mol d}}{{\text{m}}^{ - 3}}{\text{ (at 1000 K)}}$

and you have added 0.10 mol of each gasA state of matter in which a substance occupies the full volume of its container and changes shape to match the shape of the container. In a gas the distance between particles is much greater than the diameters of the particles themselves; hence the distances between particles can change as necessary so that the matter uniformly occupies its container. to a container with volume 10.0 dm3. Is the system at equilibrium? If not, will the concentration of SO3 be greater than or less than 0.010 mol dm-3 when equilibrium is reached? You can answer these questions by calculating Q and comparing it with Kc. There are three possibilities:

If Q = Kc then the actual concentrations of products (and of reactants) are equal to the equilibrium concentrations and the system is at equilibrium.

If Q < Kc then the actual concentrations of products are less than the equilibrium concentrations; the forward reaction will occur and more products will be formed.

If Q > Kc then the actual concentrations of products are greater than the equilibrium concentrations; the reverse reaction will occur and more reactants will be formed.

For the reaction given above,

$Q = \frac{{{{\{ {\text{S}}{{\text{O}}_3}\} }^2}}}{{{{\{ {\text{S}}{{\text{O}}_2}\} }^2}\{ {{\text{O}}_2}\} }} = \frac{{{{\left( {\frac{{0.10{\text{ mol}}}}{{10.0{\text{ d}}{{\text{m}}^3}}}} \right)}^2}}}{{{{\left( {\frac{{0.10{\text{ mol}}}}{{10.0{\text{ d}}{{\text{m}}^3}}}} \right)}^2}\left( {\frac{{0.10{\text{ mol}}}}{{10.0{\text{ d}}{{\text{m}}^3}}}} \right)}} = 100{\text{ mol d}}{{\text{m}}^{ - 3}}$

(In the expression for Q each actual concentration is enclosed in braces {curly brackets} in order to distinguish it from the equilibrium concentrations, which, in the Kc expression, are enclosed in [square brackets].) In this case Q = 100 mol dm-3. This is less than Kc, which has the value 245 mol dm-3. This implies that the concentrations of products are less than the equilibrium concentrations (and the concentrations of reactants are greater than the equilibrium concentrations). Therefore the reaction will proceed in the forward direction, producing more products, until the concentrations reach their equilibrium values.

EXAMPLE 1 At 2300 K, the equilibrium constantThe value of the equilibrium constant expression when equilibrium concentrations are substituted; a value greater than one indicates the position of equilibrium lies toward products (product-favored), and a value less than one indicates the position of equilibrium lies toward reactants (reactant-favored)., Kc, is 1.7 x 10-3 for the reaction
N2(g) + O2(g) $\rightleftharpoons$ 2 NO(g)
A mixtureA combination of two or more substances in which the substances retain their chemical identity. of the three gases at 2300 K has these concentrations, [N2] = 0.17 mol dm-3, [O2] = 0.17 mol dm-3, and [NO] = 0.034 mol dm-3.
(a) Is the system at equilibrium?
(b) In which direction must the reaction occur to reach equilibrium?
(c) What are the equilibrium concentrations of N2, O2, and NO?

SolutionA mixture of one or more substances dissolved in a solvent to give a homogeneous mixture. Use the known concentrations to calculate Q. Compare Q with Kc to answer questions (a) and (b). Use an ICE table to answer part (c).
$Q = \frac{{{{\left\{ {{\text{NO}}} \right\}}^2}}}{{\left\{ {{{\text{N}}_{\text{2}}}} \right\}\left\{ {{{\text{O}}_{\text{2}}}} \right\}}} = \frac{{{{\left( {0.034{\text{ mol d}}{{\text{m}}^{ - 3}}} \right)}^2}}}{{\left( {0.17{\text{ mol d}}{{\text{m}}^{ - 3}}} \right)\left( {0.17{\text{ mol d}}{{\text{m}}^{ - 3}}} \right)}} = 4.0 \times {10^{-2}}$
(a) Q is larger than Kc, so the reaction is not at equilibrium.
(b) Because Q is larger than Kc, the concentration of the product, NO, is larger than its equilibrium concentration and the concentrations of the reactants, N2 and O2, are smaller than their equilibrium concentrations. Therefore some of the product, NO, will be consumed and more of the reactants, N2 and O2, will be formed.
(c) Use the given concentrations as the initial concentrations of reactants and product. Enter these into an ICE table. Let x be the increase in the concentration of N2 as the system reacts to equilibrium. The ICE table looks like this:

 N2 O2 NO Initial concentration/mol dm-3 0.17 0.17 0.034 Change in concentration/mol dm-3 x x -2x Equilibrium concentration/mol dm-3 0.17 + x 0.17 + x 0.034 - 2x

Next, substitute the equilibrium concentrations into the Kc expression and solve for x.
${K_{\text{c}}} = 1.7 \times {10^{ - 3}} = \frac{{{{\left( {0.034 - 2x} \right)}^2}}}{{\left( {0.17 + x} \right)\left( {0.1 + x} \right)}}$
Now take the square root of both sides of this equation. This gives
$\sqrt {1.7 \times {{10}^{ - 3}}} = 0.0412 = \frac{{0.034 - 2x}}{{0.17 + x}}$
Multiplying both sides by 0.17 + x gives
0.0070 + 0.412 x = 0.034 - 2 x
2 x + 0.0412 x =0.034 - 0.0070
$x = \frac{{0.0270}}{{2.0412}} = 0.0132$
[N2] = [O2] = 0.17 + 0.013 = 0.183 mol dm-3
[NO] = 0.034 - 2(0.0132) = 0.0076mol dm-3
Check the result by substituting these concentrations into the equilibrium constant expressionThe ratio of the concentrations of products in a balanced chemical equation, each concentration raised to the power of its coefficient, divided by the concentrations of reactants, each raised to the power of its coefficient; at equilibrium this ratio has the value of the equilibrium constant..
${K_{\text{c}}} = \frac{{{{(0.0076)}^2}}}{{\left( {0.18} \right)\left( {0.18} \right)}} = 1.8 \times {10^{ - 3}}$
This agrees to two significant figures with the Kc value of 1.7 x 10-3.