Skip to content

Chemistry · Ch 8 — Physical and Chemical Equilibrium

Calculation of concentration of reactants and products at equilibrium

8.7.3

Calculation of concentration of reactants and products at equilibrium

If the equilibrium concentrations of reactants and products are known, the equilibrium constant can be calculated directly -- and, conversely, if the equilibrium constant is known, the equilibrium composition can be worked out from the starting amounts using an initial-change-equilibrium (ICE) table. Three worked cases show the method.

Formation of HI. Let 'a' mol H2H_2 and 'b' mol I2I_2 react in a container of volume V, with x mol of each consumed to form 2x mol HI:

H2H_2I2I_2HIHI
Initial molesab0
Moles reactedxx--
Moles at equilibriuma-xb-x2x

Applying the law of mass action to the equilibrium concentrations (a−xV\tfrac{a-x}{V}, b−xV\tfrac{b-x}{V}, 2xV\tfrac{2x}{V}):

KC=(2x/V)2[(a−x)/V][(b−x)/V]=4x2(a−x)(b−x)K_C = \frac{(2x/V)^2}{[(a-x)/V][(b-x)/V]} = \frac{4x^2}{(a-x)(b-x)}

Since Δng=2−2=0\Delta n_g = 2-2 = 0 for this reaction, KP=KCK_P = K_C directly.

Solved Problem. One mole of H2H_2 and one mole of I2I_2 are allowed to reach equilibrium; the equilibrium mixture contains 0.4 mol HI. So 2x=0.42x = 0.4, x=0.2x = 0.2, giving [H2]=[I2]=1−0.2=0.8[H_2]=[I_2]=1-0.2=0.8 mol and [HI]=0.4[HI]=0.4 mol (volume cancels in the ratio). KC=(0.4)2(0.8)(0.8)=0.25K_C = \dfrac{(0.4)^2}{(0.8)(0.8)} = 0.25.

Dissociation of PCl5. Let 'a' mol PCl5PCl_5 be taken in volume V, with x mol dissociating into x mol PCl3PCl_3 and x mol Cl2Cl_2:

PCl5PCl_5PCl3PCl_3Cl2Cl_2
Initial molesa00
Moles dissociatedx00
Moles at equilibriuma-xxx

KC=(x/V)(x/V)(a−x)/V=x2(a−x)VK_C = \frac{(x/V)(x/V)}{(a-x)/V} = \frac{x^2}{(a-x)V}

Since Δng=2−1=1\Delta n_g = 2-1=1, KP=KC(RT)K_P = K_C(RT). Using the ideal gas relation RT=PV/nRT = PV/n, with total equilibrium moles n=(a−x)+x+x=(a+x)n = (a-x)+x+x = (a+x), this becomes KP=x2P(a−x)(a+x)=x2Pa2−x2K_P = \dfrac{x^2P}{(a-x)(a+x)} = \dfrac{x^2P}{a^2-x^2}.

Synthesis of ammonia. Let 'a' mol N2N_2 and 'b' mol H2H_2 react in volume V, with x mol N2N_2 reacting with 3x mol H2H_2 to give 2x mol NH3NH_3:

N2N_2H2H_2NH3NH_3
Initial molesab0
Moles reactedx3x0
Moles at equilibriuma-xb-3x2x

KC=(2x/V)2[(a−x)/V][(b−3x)/V]3=4x2V2(a−x)(b−3x)3K_C = \frac{(2x/V)^2}{[(a-x)/V][(b-3x)/V]^3} = \frac{4x^2V^2}{(a-x)(b-3x)^3}

Since Δng=2−4=−2\Delta n_g = 2-4=-2, KP=KC(RT)−2K_P = K_C(RT)^{-2}; with total equilibrium moles n=a+b−2xn=a+b-2x, this expands to KP=4x2(a+b−2x)2(a−x)(b−3x)3P2K_P = \dfrac{4x^2(a+b-2x)^2}{(a-x)(b-3x)^3P^2}. …

Table 8.7.3-ice-HIICE table: formation of HI from H2 and I2
H2H_2I2I_2HIHI
Initial molesab0
Moles reactedxx--
Moles at equilibriuma-xb-x2x
Misc 8.7.3-derive-HIDerived Kc and Kp for HI formation

Worked out. Applying the law of mass action to the ICE table above: KC=(2x/V)2[(a−x)/V][(b−x)/V]=4x2(a−x)(b−x)K_C = \dfrac{(2x/V)^2}{[(a-x)/V][(b-x)/V]} = \dfrac{4x^2}{(a-x)(b-x)}. Since Δng=2−2=0\Delta n_g = 2 - 2 = 0 for this reaction, KP=KCK_P = K_C. …

Misc 8.7.3-solved-HISolved Problem: Kc for HI formation from 1 mol H2 + 1 mol I2

Worked out. 1 mol H2_2 and 1 mol I2_2 react; at equilibrium the mixture contains 0.4 mol HI, so 2x=0.42x=0.4, x=0.2x=0.2, giving [H2]=[I2]=0.8[H_2]=[I_2]=0.8 mol and [HI]=0.4[HI]=0.4 mol (same volume cancels). KC=(0.4)2(0.8)(0.8)=0.25K_C = \dfrac{(0.4)^2}{(0.8)(0.8)} = 0.25. …

Table 8.7.3-ice-PCl5ICE table: dissociation of PCl5
PCl5PCl_5PCl3PCl_3Cl2Cl_2
Initial molesa00
Moles dissociatedx00
Moles at equilibriuma-xxx
Misc 8.7.3-derive-PCl5Derived Kc and Kp for PCl5 dissociation

Worked out. KC=(x/V)2(a−x)/V=x2(a−x)VK_C = \dfrac{(x/V)^2}{(a-x)/V} = \dfrac{x^2}{(a-x)V}. Since Δng=2−1=1\Delta n_g = 2-1 = 1, KP=KC(RT)K_P = K_C(RT); using RT=PV/nRT = PV/n with total moles n=(a+x)n=(a+x) at equilibrium, this rearranges to KP=x2P(a−x)(a+x)=x2Pa2−x2K_P = \dfrac{x^2P}{(a-x)(a+x)} = \dfrac{x^2P}{a^2-x^2}. …

Table 8.7.3-ice-NH3ICE table: synthesis of ammonia
N2N_2H2H_2NH3NH_3
Initial molesab0
Moles reactedx3x0
Moles at equilibriuma-xb-3x2x
Misc 8.7.3-derive-NH3Derived Kc and Kp for ammonia synthesis

Worked out. KC=(2x/V)2[(a−x)/V][(b−3x)/V]3=4x2V2(a−x)(b−3x)3K_C = \dfrac{(2x/V)^2}{[(a-x)/V][(b-3x)/V]^3} = \dfrac{4x^2V^2}{(a-x)(b-3x)^3}. Since Δng=2−4=−2\Delta n_g = 2-4=-2, KP=KC(RT)−2K_P = K_C(RT)^{-2}; with total equilibrium moles n=a+b−2xn=a+b-2x, this becomes KP=4x2(a+b−2x)2(a−x)(b−3x)3P2K_P = \dfrac{4x^2(a+b-2x)^2}{(a-x)(b-3x)^3P^2}. …

Misc 8.7.3-solved-1Solved Problem 1: Kc for ammonia formation from given equilibrium concentrations

Worked out. [NH3]=1.8×10−2[NH_3] = 1.8\times10^{-2} M, [N2]=1.2×10−2[N_2] = 1.2\times10^{-2} M, [H2]=3×10−2[H_2] = 3\times10^{-2} M. KC=[NH3]2[N2][H2]3=(1.8×10−2)2(1.2×10−2)(3×10−2)3≈1×103K_C = \dfrac{[NH_3]^2}{[N_2][H_2]^3} = \dfrac{(1.8\times10^{-2})^2}{(1.2\times10^{-2})(3\times10^{-2})^3} \approx 1\times10^3 L2^2 mol−2^{-2}. …

Misc 8.7.3-solved-2Solved Problem 2: equilibrium concentration of D from Kc

Worked out. A+B⇌C+DA + B \rightleftharpoons C + D, KC=100K_C = 100 at 298 K, all four initial concentrations 1 M. With x mol reacted, equilibrium gives [A]=[B]=1−x[A]=[B]=1-x, [C]=[D]=1+x[C]=[D]=1+x. KC=(1+x)2(1−x)2=100⇒1+x1−x=10⇒x=0.818K_C = \dfrac{(1+x)^2}{(1-x)^2}=100 \Rightarrow \dfrac{1+x}{1-x}=10 \Rightarrow x = 0.818. [D]eq=1+x=1.818[D]_{eq} = 1+x = 1.818 M. …

Misc 8.7.3-evaluate-yourselfEvaluate Yourself: PCl5 equilibrium composition

Worked out. 1 mol PCl5 kept in a closed 1 dm3^3 container is allowed to reach equilibrium at 423 K, where KC=2K_C=2 for the dissociation. Calculate the equilibrium composition of the reaction mixture. …

Start your 14-day free trial to unlock the full solution →
← Previous · 8.7.2Predicting the direction of a reactionNext · 8.8 →Le-Chatelier's Principle