
For the reaction at $800K$,
${N_{2(g)}} + 3{H_{2(g)}} \rightleftharpoons 2N{H_{3(g)}}$
The ratio of ${K_p}$ and ${K_c}$ is ($R = 0.082Latm/molK$ ) :
A. $2.3 \times {10^{ - 4}}$
B. $3.2 \times {10^{ - 6}}$
C. $2.3 \times {10^4}$
D. $3.2 \times {10^6}$
Answer
565.8k+ views
Hint: The equilibrium constant of a chemical reaction is the value of its reaction quotient at chemical equilibrium, a state approached by a dynamic chemical system after sufficient time has elapsed at which its composition has no measurable tendency towards further change. For a given set of reaction conditions, the equilibrium constant is independent of the initial analytical concentrations of the reactant and product species in the mixture.
Complete answer:
The relation between equilibrium constant at constant pressure and equilibrium constant at constant concentration is as follows:
${K_p} = {K_c}{(RT)^{\Delta {n_g}}}$ … (i)
Where, ${K_p} = $ equilibrium constant at constant pressure
${K_c} = $ equilibrium constant at constant concentration
$R = $ universal gaseous constant $ = 0.082Latm/molK$
$T = $ temperature of the reaction$ = 800K$
The term $\Delta {n_g}$ plays a crucial role in determining the position of the equilibrium constant and the value of reaction quotient which determines the shifting of equilibrium.
$\Delta {n_g} = {n_p} - {n_r}$ = Difference in the number of gaseous moles of the products and the number of gaseous moles of the reactants
In this case, $\Delta {n_g} = 2 - 4 = - 2$
Substituting these values in equation (i), we have:
$ \Rightarrow \dfrac{{{K_p}}}{{{K_c}}} = {(RT)^{ - 2}}$
Thus, on solving, we have:
$ \Rightarrow \dfrac{{{K_p}}}{{{K_c}}} = \dfrac{1}{{{{(0.0821 \times 800)}^2}}}$
$ \Rightarrow \dfrac{{{K_p}}}{{{K_c}}} = \dfrac{1}{{4313.8624}} = 2.3 \times {10^{ - 4}}$
The ratio of the equilibrium constant at constant pressure and equilibrium constant at constant concentration is equal to $2.3 \times {10^{ - 4}}$ .
Thus option A is the correct answer.
Note:
The pressure dependence of the equilibrium constant is usually weak in the range of pressures normally encountered in industry, and therefore, it is usually neglected in practice. This is true for condensed reactants/products (i.e., when reactants and products are solids or liquid) as well as gaseous ones.
Complete answer:
The relation between equilibrium constant at constant pressure and equilibrium constant at constant concentration is as follows:
${K_p} = {K_c}{(RT)^{\Delta {n_g}}}$ … (i)
Where, ${K_p} = $ equilibrium constant at constant pressure
${K_c} = $ equilibrium constant at constant concentration
$R = $ universal gaseous constant $ = 0.082Latm/molK$
$T = $ temperature of the reaction$ = 800K$
The term $\Delta {n_g}$ plays a crucial role in determining the position of the equilibrium constant and the value of reaction quotient which determines the shifting of equilibrium.
$\Delta {n_g} = {n_p} - {n_r}$ = Difference in the number of gaseous moles of the products and the number of gaseous moles of the reactants
In this case, $\Delta {n_g} = 2 - 4 = - 2$
Substituting these values in equation (i), we have:
$ \Rightarrow \dfrac{{{K_p}}}{{{K_c}}} = {(RT)^{ - 2}}$
Thus, on solving, we have:
$ \Rightarrow \dfrac{{{K_p}}}{{{K_c}}} = \dfrac{1}{{{{(0.0821 \times 800)}^2}}}$
$ \Rightarrow \dfrac{{{K_p}}}{{{K_c}}} = \dfrac{1}{{4313.8624}} = 2.3 \times {10^{ - 4}}$
The ratio of the equilibrium constant at constant pressure and equilibrium constant at constant concentration is equal to $2.3 \times {10^{ - 4}}$ .
Thus option A is the correct answer.
Note:
The pressure dependence of the equilibrium constant is usually weak in the range of pressures normally encountered in industry, and therefore, it is usually neglected in practice. This is true for condensed reactants/products (i.e., when reactants and products are solids or liquid) as well as gaseous ones.
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