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Sagot :
Dynamic equilibrium in a chemical reaction is a state where the rates of the forward and reverse reactions are equal. At this point, the concentration of reactants and products remains constant over time, though they are not necessarily equal. This occurs because the rate at which the reactants form products is equal to the rate at which the products revert to reactants.
Let's analyze the options given in the context of the reaction:
[tex]\[ 2 H_2 + O_2 \rightleftarrows 2 H_2O \][/tex]
A. There will be no more [tex]$H _2$[/tex] or [tex]$O _2$[/tex].
- This is incorrect. At dynamic equilibrium, both reactants ([tex]$H_2$[/tex] and [tex]$O_2$[/tex]) and products ([tex]$H_2O$[/tex]) are present. The reaction doesn't go to completion in either direction, so there will always be some [tex]$H_2$[/tex] and [tex]$O_2$[/tex] left.
B. The pressure will reach its minimum.
- This is not necessarily true. The pressure in a closed system depends on the total number of moles of gas and the volume and temperature of the system. While the pressure may change as the reaction proceeds toward equilibrium, it does not imply it will reach its minimum value at equilibrium.
C. [tex]$H_2O$[/tex], [tex]$H_2$[/tex], and [tex]$O_2$[/tex] will all form at the same rate.
- This is correct. At dynamic equilibrium, the rate at which [tex]$H_2O$[/tex] forms from [tex]$H_2$[/tex] and [tex]$O_2$[/tex] is equal to the rate at which [tex]$H_2$[/tex] and [tex]$O_2$[/tex] form from [tex]$H_2O$[/tex]. Hence, [tex]$H_2O$[/tex], [tex]$H_2$[/tex], and [tex]$O_2$[/tex] will all form at the same rate.
D. The concentration of [tex]$H_2O$[/tex] will reach its maximum.
- This is incorrect. The concentration of [tex]$H_2O$[/tex] will not necessarily reach its maximum at equilibrium. Instead, it will reach a constant value that depends on the initial concentrations and the equilibrium constant of the reaction.
Thus, the correct answer is:
C. [tex]\(\mathbf{H_{2}O \, , \, H_{2} \, , \, and\, O_{2}\, will\, all\, form\, at\, the\, same\, rate.} \)[/tex]
Let's analyze the options given in the context of the reaction:
[tex]\[ 2 H_2 + O_2 \rightleftarrows 2 H_2O \][/tex]
A. There will be no more [tex]$H _2$[/tex] or [tex]$O _2$[/tex].
- This is incorrect. At dynamic equilibrium, both reactants ([tex]$H_2$[/tex] and [tex]$O_2$[/tex]) and products ([tex]$H_2O$[/tex]) are present. The reaction doesn't go to completion in either direction, so there will always be some [tex]$H_2$[/tex] and [tex]$O_2$[/tex] left.
B. The pressure will reach its minimum.
- This is not necessarily true. The pressure in a closed system depends on the total number of moles of gas and the volume and temperature of the system. While the pressure may change as the reaction proceeds toward equilibrium, it does not imply it will reach its minimum value at equilibrium.
C. [tex]$H_2O$[/tex], [tex]$H_2$[/tex], and [tex]$O_2$[/tex] will all form at the same rate.
- This is correct. At dynamic equilibrium, the rate at which [tex]$H_2O$[/tex] forms from [tex]$H_2$[/tex] and [tex]$O_2$[/tex] is equal to the rate at which [tex]$H_2$[/tex] and [tex]$O_2$[/tex] form from [tex]$H_2O$[/tex]. Hence, [tex]$H_2O$[/tex], [tex]$H_2$[/tex], and [tex]$O_2$[/tex] will all form at the same rate.
D. The concentration of [tex]$H_2O$[/tex] will reach its maximum.
- This is incorrect. The concentration of [tex]$H_2O$[/tex] will not necessarily reach its maximum at equilibrium. Instead, it will reach a constant value that depends on the initial concentrations and the equilibrium constant of the reaction.
Thus, the correct answer is:
C. [tex]\(\mathbf{H_{2}O \, , \, H_{2} \, , \, and\, O_{2}\, will\, all\, form\, at\, the\, same\, rate.} \)[/tex]
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