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Standard XI

Chemistry

Derivation of Kp and Kc

Kc for the re...

Question

$K_{c}$ for the reaction $N_{2} O_{4} ⇌ 2 N O_{2}$ in chloroform at 291 K is 1.14. Calculate the $K_{p}$ of the reaction when the concentration of the two gases are $0.5 m o l d m^{- 3}$ each at the same temperature. $(R = 0.082 L a t m K^{- 1} m o l^{- 1})$

-27.1

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1.1

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27.1

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Solution

The correct option is D 27.1
From the given data
$T = 291 K; R = 0.082 l i t a t m K^{- 1} m o l^{- 1}$
$K_{c} = 1.14; c = 0.5 m o l d m^{- 3}$
$△ n = 2 - 1 = 1$ in this case
we know the relation,
$K_{p} = K_{c} (R T)^{△ n}$
$K_{p} = 1.14 (0.082 \times 291) = 27.1$

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Similar questions

K_{c}

for the reaction

N_{2} O_{4} ⇌ 2 N O_{2}

in chloroform at 291 K is 1.14. Calculate the

K_{p}

of the reaction when the concentration of the two gases are

0.5 m o l d m^{- 3}

each at the same temperature.

(R = 0.082 L a t m K^{- 1} m o l^{- 1})

Q. Standard Free Energy and Equilibrium constant: The change in free energy for a reaction taking place between gaseous reactants and products represented by general equation.

Δ G = Δ G^{\circ} + R T l n Q_{P}

the condition for a system to be at equilibrium is that

Δ G = 0 and Q_{p} = K_{P}

Thus at equilibrium

Δ G^{\circ} = - R T l n K_{P}

Note: In the reaction, where all gaseous reactants and products;

K represents K_{P}

, we may conclude that for standard reactions, i.e., at

1 M or 1 atm

When

Δ G^{\circ} = - v e or K > 1 : forward reaction is feasible

Δ G^{\circ} = + v e or K < 1 : reverse reaction is feasible

Δ G^{\circ} = 0 or K = 1 : reaction is at equilibrium (very rare)

K_{c}

for reaction

N_{2} O_{4} ⇌ 2 N O_{2}

in chloroform at

291 K

1.14

. Calculate the free energy change of the reaction when the concentration of the two gases are

0.5 {mol dm}^{- 3}

each at the same temperature. (

R = 0.082 {lit atm K}^{- 1} {mol}^{- 1}

)

Q. For the equilibrium

A_{(g)} ⇌ B_{(g)} + C_{(g)}; K_{p} = 0.82

atm at

27^{o}

C. At the same temperature, its

K_{c}

in mol

l i t^{- 1}

is:

[R = 0.082

lit atm

m o l^{- 1} K^{- 1}]

Q. At 700 K, the equilibrium constant

K_{p},

for the reaction

2 S O_{3} (g) ⇌ 2 S O_{2} (g) + O_{2} (g)

1.8 \times 10^{- 3} k P a .

What is the numerical value of

K_{C}

for this reaction at the same temperature?

R = 0.0821 L a t m m o l^{- 1} K^{- 1}

Q. For the reaction

N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g); △ H = - 46 k J m o l^{- 1}

Calculate the value of

K_{p} .

Given

K_{C} = 0.5 l i t^{2} m o l^{- 2}

and

R = 0.082 L a t m m o l^{- 1} K^{- 1}

400^{o} C

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Derivation of Kp and Kc

Standard XI Chemistry

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Kc for the reaction N2O4⇌2NO2 in chloroform at 291 K is 1.14. Calculate the Kp of the reaction when the concentration of the two gases are 0.5 mol dm−3 each at the same temperature. (R=0.082 L atm K−1 mol−1)

The correct option is D 27.1From the given data T=291 K;R=0.082 lit atm K−1 mol−1 Kc=1.14;c=0.5 mol dm−3 △n=2−1=1 in this case we know the relation, Kp=Kc(RT)△n Kp=1.14(0.082×291)=27.1

$K_{c}$ for the reaction $N_{2} O_{4} ⇌ 2 N O_{2}$ in chloroform at 291 K is 1.14. Calculate the $K_{p}$ of the reaction when the concentration of the two gases are $0.5 m o l d m^{- 3}$ each at the same temperature. $(R = 0.082 L a t m K^{- 1} m o l^{- 1})$

The correct option is D 27.1
From the given data
$T = 291 K; R = 0.082 l i t a t m K^{- 1} m o l^{- 1}$
$K_{c} = 1.14; c = 0.5 m o l d m^{- 3}$
$△ n = 2 - 1 = 1$ in this case
we know the relation,
$K_{p} = K_{c} (R T)^{△ n}$
$K_{p} = 1.14 (0.082 \times 291) = 27.1$