The effect of a change in pressure
The expression given below is known as the reaction quotient. The form of Q depends on the equation for the reaction. When the system is at equilibrium Q equals K, the equilibrium constant for the reaction.
For a system at equilibrium (Q = K) at constant temperature Q becomes unequal to K for a reaction involving gases, if
If the total amount in moles of gaseous reactants differs from the total amount in moles of gaseous products
If aA + bB 
cC + dD
cC + dDand A, B, C and D are substances that are all dissolved (solutes) or are all gases
where a, b, c and d are coefficients in the balanced equation
and [A] = concentration of A in
where a, b, c and d are coefficients in the balanced equation
and [A] = concentration of A in
| Q = | [C]c[D]d |
| [A]a[B]b |
For a system at equilibrium (Q = K) at constant temperature Q becomes unequal to K for a reaction involving gases, if
the volume of the reaction vessel is changed (results in a change in all concentrations) and
the number of moles of gaseous reactants in the equation for the reaction differs from the number of moles of gaseous products
Reaction 1: 2NO2(g) 
N2O4(g) (2 moles gas at left ,1 at right - change in volume makes Q ≠ K)
Reaction 2: H2(g) +Cl2(g) 2HCl(g) (2 moles gas at left, 2 at right - change volume Q still equals K)
N2O4(g) (2 moles gas at left ,1 at right - change in volume makes Q ≠ K)Reaction 2: H2(g) +Cl2(g)
2HCl(g) (2 moles gas at left, 2 at right - change volume Q still equals K)Show/Hide why this is true
Consider the Q expression where the concentrations are expressed as n/V.
For example, if all concentrations are doubled and
if a + b = c + d, the two volume expressions cancel out, and Q does NOT change when the volume changes.
If a + b > c + d, the sum of the powers of concentrations for reactants is greater than for products.
Increasing V (diluting the mixture or decreasing the pressure) makes Q greater than K because Va+b is in the numerator, and reaction occurs in the reverse direction.
If a + b < c + d, the sum of the powers of the concentrations for products is greater than reactants.
Increasing V (diluting the misture or decreasing the pressure) makes Q less than K because Vc+d is in the denominator, and reaction occurs in the forward direction
| Q = | n(C)c × n(D)d × Va+b |
| n(A)a × n(B)b × Vc+d |
For example, if all concentrations are doubled and
if a + b = c + d, the two volume expressions cancel out, and Q does NOT change when the volume changes.
If a + b > c + d, the sum of the powers of concentrations for reactants is greater than for products.
Increasing V (diluting the mixture or decreasing the pressure) makes Q greater than K because Va+b is in the numerator, and reaction occurs in the reverse direction.
If a + b < c + d, the sum of the powers of the concentrations for products is greater than reactants.
Increasing V (diluting the misture or decreasing the pressure) makes Q less than K because Vc+d is in the denominator, and reaction occurs in the forward direction
If the total amount in moles of gaseous reactants differs from the total amount in moles of gaseous products
and pressure is increased (volume decreased),
equilibrium is restored by reaction to decrease the total amount in moles of gas present.
For Reaction 1, increase P (decrease V), after equilibrium restored, [NO2] higher (3 → 2 mole).
and pressure is decreased (volume increased),
equilibrium is restored by reaction to increase in the total amount in moles of gas present.
equilibrium is restored by reaction to decrease the total amount in moles of gas present.
For Reaction 1, increase P (decrease V), after equilibrium restored, [NO2] higher (3 → 2 mole).
and pressure is decreased (volume increased),
equilibrium is restored by reaction to increase in the total amount in moles of gas present.
For Reaction 1, decrease P (increase V), after equilibrium restored, [NO] and [O2] higher (2→3 mole)