(C) Given reaction: $NH_{3(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)}$ with $K_{c1} = 4$.For the target reaction: $N_{2(g)} +

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$\left( \frac{1}{4 \times 800R} \right)^2$

(C) Given reaction: $NH_{3(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)}$ with $K_{c1} = 4$.For the target reaction: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$,we reverse the first reaction and multiply by $2$.Thus,$K_{c2} = (1/K_{c1})^2 = (1/4)^2 = 1/16$.The relationship between $K_P$ and $K_c$ is $K_P = K_c(RT)^{\Delta n}$.For the target reaction,$\Delta n = 2 - (1 + 3) = -2$.Temperature $T = 527 + 273 = 800 \ K$.$K_P = (1/16) \times (R \times 800)^{-2} = (1/4)^2 \times (800R)^{-2} = \left( \frac{1}{4 \times 800R} \right)^2$.

Class 11 Chemistry 6-1.Equilibrium (Chemical Equilibrium)Kp and Kc Relationship

Medium

At $527 \ ^oC$,the reaction given below has $K_c = 4$.
$NH_{3(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)}$
What is the $K_P$ for the following reaction?
$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$

A
$16 \times (800 \ R)^2$
B
$\left( \frac{800R}{4} \right)^{-2}$
C
$\left( \frac{1}{4 \times 800R} \right)^2$
D
None of these

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6-1.Equilibrium (Chemical Equilibrium)

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Kp and Kc Relationship

If $2 \ mol$ of $H_2$ and $I_2$ are taken initially in a $1 \ L$ vessel,and the equilibrium concentration of $HI$ is $2 \ mol/L$,find the $K_p$ for the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$.

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For the reaction $A_{(g)} + 2B_{(g)} \rightleftharpoons 2C_{(g)}$,$1 \ mol$ of $A$ and $1.5 \ mol$ of $B$ are taken in a $2 \ L$ vessel. If the concentration of $C$ at equilibrium is $0.35 \ M$,then the equilibrium constant $K_c$ of the reaction will be ....... $M^{-1}$.

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For the reaction $A + B \rightleftharpoons C + D$ at $298 \ K$,the equilibrium constant is $10.0$. If the initial concentration of all four species is $1 \ M$,what will be the equilibrium concentration of $D$ (in $mol \ L^{-1}$)?

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At $800 \ K$ in a closed vessel,the molar concentrations of $N_2, O_2$ and $NO$ at equilibrium are $3.2 \times 10^{-3} \ M, 4.2 \times 10^{-3} \ M$ and $2.8 \times 10^{-3} \ M$ respectively. The approximate values of $K_{c}$ and $\frac{1}{K_{c}}$ for the following reaction are respectively: $N_{2(g)} + O_{2(g)} \rightleftharpoons 2 NO_{(g)}$

MediumAP EAMCET 2017

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The equilibrium constant for the reversible reaction,$N_2 + 3H_2 \rightleftharpoons 2NH_3$ is $K$ and for the reaction $\frac{1}{2}N_2 + \frac{3}{2}H_2 \rightleftharpoons NH_3$ the equilibrium constant is $K'$. $K$ and $K'$ will be related as

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At $527 \ ^oC$,the reaction given below has $K_c = 4$.$NH_{3(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)}$What is the $K_P$ for the following reaction?$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$

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If $2 \ mol$ of $H_2$ and $I_2$ are taken initially in a $1 \ L$ vessel,and the equilibrium concentration of $HI$ is $2 \ mol/L$,find the $K_p$ for the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$.

For the reaction $A_{(g)} + 2B_{(g)} \rightleftharpoons 2C_{(g)}$,$1 \ mol$ of $A$ and $1.5 \ mol$ of $B$ are taken in a $2 \ L$ vessel. If the concentration of $C$ at equilibrium is $0.35 \ M$,then the equilibrium constant $K_c$ of the reaction will be ....... $M^{-1}$.

For the reaction $A + B \rightleftharpoons C + D$ at $298 \ K$,the equilibrium constant is $10.0$. If the initial concentration of all four species is $1 \ M$,what will be the equilibrium concentration of $D$ (in $mol \ L^{-1}$)?

The equilibrium constant for the reversible reaction,$N_2 + 3H_2 \rightleftharpoons 2NH_3$ is $K$ and for the reaction $\frac{1}{2}N_2 + \frac{3}{2}H_2 \rightleftharpoons NH_3$ the equilibrium constant is $K'$. $K$ and $K'$ will be related as

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At $527 \ ^oC$,the reaction given below has $K_c = 4$.
$NH_{3(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)}$
What is the $K_P$ for the following reaction?
$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$