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The reaction can proceed in both the forward and reverse directions.


When a system has reached equilibrium, no further changes in the reactant and product concentrations occur; the forward and reverse reactions continue to proceed, but at equal rates.


Not necessarily. A system at equilibrium is characterized by constant reactant and product concentrations, but the values of the reactant and product concentrations themselves need not be equal.


Equilibrium cannot be established between the liquid and the gas phase if the top is removed from the bottle because the system is not closed; one of the components of the equilibrium, the Br2 vapor, would escape from the bottle until all liquid disappeared. Thus, more liquid would evaporate than can condense back from the gas phase to the liquid phase.


(a) Kc = [Ag+][Cl] < 1. AgCl is insoluble; thus, the concentrations of ions are much less than 1 M; (b) Kc=1[Pb2+][Cl]2Kc=1[Pb2+][Cl]2 > 1 because PbCl2 is insoluble and formation of the solid will reduce the concentration of ions to a low level (<1 M).


Since Kc=[C6H6][C2H2]3,Kc=[C6H6][C2H2]3, a value of Kc ≈ 10 means that C6H6 predominates over C2H2. In such a case, the reaction would be commercially feasible if the rate to equilibrium is suitable.


Kc > 1


(a) Qc=[CH3Cl][HCl][CH4][Cl2];Qc=[CH3Cl][HCl][CH4][Cl2]; (b) Qc=[NO]2[N2][O2];Qc=[NO]2[N2][O2]; (c) Qc=[SO3]2[SO2]2[O2];Qc=[SO3]2[SO2]2[O2]; (d) Qc = [SO2]; (e) Qc=1[P4][O2]5;Qc=1[P4][O2]5; (f) Qc=[Br]2[Br2];Qc=[Br]2[Br2]; (g) Qc=[CO2][CH4][O2]2;Qc=[CO2][CH4][O2]2; (h) Qc = [H2O]5


(a) Qc 25 proceeds left; (b) QP 0.22 proceeds right; (c) Qc undefined proceeds left; (d) QP 1.00 proceeds right; (e) QP 0 proceeds right; (f) Qc 4 proceeds left


The system will shift toward the reactants to reach equilibrium.


(a) homogenous; (b) homogenous; (c) homogenous; (d) heterogeneous; (e) heterogeneous; (f) homogenous; (g) heterogeneous; (h) heterogeneous


This situation occurs in (a) and (b).


(a) KP = 1.6 ×× 10−4; (b) KP = 50.2; (c) Kc = 5.31 ×× 10−39; (d) Kc = 4.60 ×× 10−3


K P = P H 2 O = 0.042 . K P = P H 2 O = 0.042 .


Q c = [ NH 4 + ] [ OH ] [ NH 3 ] Q c = [ NH 4 + ] [ OH ] [ NH 3 ]


The amount of CaCO3 must be so small that PCO2PCO2 is less than KP when the CaCO3 has completely decomposed. In other words, the starting amount of CaCO3 cannot completely generate the full PCO2PCO2 required for equilibrium.


The change in enthalpy may be used. If the reaction is exothermic, the heat produced can be thought of as a product. If the reaction is endothermic the heat added can be thought of as a reactant. Additional heat would shift an exothermic reaction back to the reactants but would shift an endothermic reaction to the products. Cooling an exothermic reaction causes the reaction to shift toward the product side; cooling an endothermic reaction would cause it to shift to the reactants’ side.


No, it is not at equilibrium. Because the system is not confined, products continuously escape from the region of the flame; reactants are also added continuously from the burner and surrounding atmosphere.


Add N2; add H2; decrease the container volume; heat the mixture.


(a) T increase = shift right, V decrease = shift left; (b) T increase = shift right, V = no effect; (c) T increase = shift left, V decrease = shift left; (d) T increase = shift left, V decrease = shift right.


(a) Kc=[CH3OH][H2]2[CO];Kc=[CH3OH][H2]2[CO]; (b) [H2] increases, [CO] decreases, [CH3OH] increases; (c), [H2] increases, [CO] decreases, [CH3OH] decreases; (d), [H2] increases, [CO] increases, [CH3OH] increases; (e), [H2] increases, [CO] increases, [CH3OH] decreases


(a) Kc=[CO][H2][H2O];Kc=[CO][H2][H2O]; (b) [H2O] no change, [CO] no change, [H2] no change; (c) [H2O] decreases, [CO] decreases, [H2] decreases; (d) [H2O] increases, [CO] increases, [H2] decreases; (f) [H2O] decreases, [CO] increases, [H2] increases. In (b), (c), (d), and (e), the mass of carbon will change, but its concentration (activity) will not change.


Only (b)


Add NaCl or some other salt that produces Cl to the solution. Cooling the solution forces the equilibrium to the right, precipitating more AgCl(s).


Though the solution is saturated, the dynamic nature of the solubility equilibrium means the opposing processes of solid dissolution and precipitation continue to occur (just at equal rates, meaning the dissolved ion concentrations and the amount of undissolved solid remain constant). The radioactive Ag+ ions detected in the solution phase come from dissolution of the added solid, and their presence is countered by precipitation of nonradioactive Ag+.


Kc=[C]2[A][B]2.Kc=[C]2[A][B]2. [A] = 0.1 M, [B] = 0.1 M, [C] = 1 M; and [A] = 0.01, [B] = 0.250, [C] = 0.791.


Kc = 6.00 ×× 10−2


Kc = 0.50


KP = 1.9 ×× 103


KP = 3.06


(a) −2x, +2x; (b) +43x+43x, 23x23x, −2x; (c) −2x, 3x; (d) +x, −x, −3x; (e) +x; (f) 14x14x.


Activities of pure crystalline solids equal 1 and are constant; however, the mass of Ni does change.


[NH3] = 9.1 ×× 10−2 M


PBrCl = 4.9 ×× 10−2 atm


[CO] = 2.04 ×× 10−4 M


P H 2 O = 3.64 × 10 −3 atm P H 2 O = 3.64 × 10 −3 atm


Calculate Q based on the calculated concentrations and see if it is equal to Kc. Because Q does equal 4.32, the system must be at equilibrium.


(a) [NO2] = 1.17 ×× 10−3 M
[N2O4] = 0.128 M
(b) The assumption that x is negligibly small compared to 0.129 is confirmed by comparing the initial concentration of the N2O4 to its concentration at equilibrium (they differ by just 1 in the least significant digit’s place).


(a) [H2S] = 0.810 atm
[H2] = 0.014 atm
[S2] = 0.0072 atm
(b) The assumption that 2x is negligibly small compared to 0.824 is confirmed by comparing the initial concentration of the H2S to its concentration at equilibrium (0.824 atm versus 0.810 atm, a difference of less than 2%).


507 g


330 g


(a) 0.33 mol.
(b) [CO]2 = 0.50 M. Added H2 forms some water as a result of a shift to the left after H2 is added.


(a) Kc=[NH3]4[O2]7[NO2]4[H2O]6.Kc=[NH3]4[O2]7[NO2]4[H2O]6. (b) [NH3] must increase for Qc to reach Kc. (c) The increase in system volume would lower the partial pressures of all reactants (including NO2). (d) PO2=49torrPO2=49torr


P N 2 O 3 = 1.90 atm and P NO = P NO 2 = 1.90 atm P N 2 O 3 = 1.90 atm and P NO = P NO 2 = 1.90 atm


In each of the following, the value of ΔG is not given at the temperature of the reaction. Therefore, we must calculate ΔG from the values ΔH° and ΔS and then calculate ΔG from the relation ΔG = ΔH° − TΔS°.
(a) K = 1.07 × 10–13;
(b) K = 2.42 × 10−3;
(c) K = 2.73 × 104;
(d) K = 0.229;
(e) K = 16.1


The standard free energy change is ΔG298°=RTlnK=4.84 kJ/mol.ΔG298°=RTlnK=4.84 kJ/mol. When reactants and products are in their standard states (1 bar or 1 atm), Q = 1. As the reaction proceeds toward equilibrium, the reaction shifts left (the amount of products drops while the amount of reactants increases): Q < 1, and ΔG298ΔG298 becomes less positive as it approaches zero. At equilibrium, Q = K, and ΔG = 0.


The reaction will be spontaneous at temperatures greater than 287 K.


K = 5.35 ×× 1015
The process is exothermic.


1.0 ×× 10−8 atm. This is the maximum pressure of the gases under the stated conditions.


x = 1.29 × 10 5 atm = P O 2 x = 1.29 × 10 5 atm = P O 2


−0.16 kJ

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