The equilibrium constant for the reaction H2+ (1)/(2) O2 ⇌H2 O at 1 atm and 1500^∘ C is given to

step 1 Problem number 111. This reaction is a double reaction. So option B is the only one with a diffe

The equilibrium constant for the reaction H2+ (1)/(2) O2 ⇌H2...

The equilibrium constant for the reaction $\mathrm{H}_{2}+$ $\frac{1}{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{H}_{2} \mathrm{O}$ at 1 atm and $1500^{\circ} \mathrm{C}$ is given to be $K .$ Of the reactions given below, all at $1500^{\circ} \mathrm{C}$, the reaction that has a different equilibrium constant is $(a) \mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{H}_{2} \mathrm{O}$ at 5 atm $(b) 2 \mathrm{H}_{2}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}$ at 1 atm $(c) \mathrm{H}_{2}+\mathrm{O}_{2} \rightleftharpoons \mathrm{H}_{2} \mathrm{O}+\frac{1}{2} \mathrm{O}_{2}$ at 2 atm $(d) \mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2}+3 \mathrm{N}_{2} \rightleftharpoons \mathrm{H}_{2} \mathrm{O}+3 \mathrm{N}_{2} $ at $5 \mathrm{atm}$ $(e) \mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2}+3 \mathrm{N}_{2} \rightleftharpoons \mathrm{H}_{2} \mathrm{O}+3 \mathrm{N}_{2} \quad$ at 1 atm

The equilibrium constant for the reaction $\mathrm{H}_{2}+$ $\frac{1}{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{H}_{2} \mathrm{O}$ at 1 atm and $1500^{\circ} \mathrm{C}$ is given to be $K .$ Of the reactions given below, all at $1500^{\circ} \mathrm{C}$, the reaction that has a different equilibrium constant is
$(a) \mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2} \rightleftharpoons \mathrm{H}_{2} \mathrm{O}$ at 5 atm
$(b) 2 \mathrm{H}_{2}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}$ at 1 atm
$(c) \mathrm{H}_{2}+\mathrm{O}_{2} \rightleftharpoons \mathrm{H}_{2} \mathrm{O}+\frac{1}{2} \mathrm{O}_{2}$ at 2 atm
$(d) \mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2}+3 \mathrm{N}_{2} \rightleftharpoons \mathrm{H}_{2} \mathrm{O}+3 \mathrm{N}_{2} $ at $5 \mathrm{atm}$
$(e) \mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2}+3 \mathrm{N}_{2} \rightleftharpoons \mathrm{H}_{2} \mathrm{O}+3 \mathrm{N}_{2} \quad$ at 1 atm

Key Concepts

Equilibrium Constant

The equilibrium constant (K) is defined as the ratio of the activities or effective concentrations (or partial pressures, when expressed relative to a standard state) of the products to that of the reactants at equilibrium. It is a fundamental measure of the relative extent to which a reaction proceeds and is uniquely determined by the temperature, regardless of the overall pressure, as long as the reaction is defined in terms of activities.

Reaction Stoichiometry and Its Impact on K

The numerical value of an equilibrium constant depends on the form of the balanced chemical equation. When the reaction equation is manipulated by changing its stoichiometric coefficients, the equilibrium constant is modified accordingly. For example, if the overall reaction is multiplied by a factor, the new equilibrium constant becomes the original K raised to that power, highlighting the sensitivity of K to stoichiometric changes.

Pressure Effects on Equilibrium Constants

For gas-phase reactions, while the equilibrium position (in terms of partial pressures) may shift when the total pressure changes, the equilibrium constant expressed in terms of activities (or standardized partial pressures) remains constant at a given temperature. Thus, altering the total pressure of the system does not change K as long as the same reference state is used.

Role of Inert Gases in Equilibrium Systems

Inert gases, which do not participate directly in the chemical reaction, typically do not affect the equilibrium constant. They may change the total pressure of the system, but since they do not contribute to the reaction quotient, the equilibrium position in terms of the reacting species remains unaffected when expressed in standardized units.

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