At 700 K- the equilibrium constant Kp- for the reaction 2SO3g - 2SO2g - O2gis 1-8 - 10-3 kPa- What is the numerical value of KC for this reaction at the same temperature-R-0-0821 L atm mol-1 K-1

We know the relationship nbsp; K_P = K_C(RT)^ n Here K_P = 1.80 × 10^ 3 K_P = 1.8 × 10^ 3101.3 atm = 1.78 × 10^ 5 atm nbsp; R = 0.0821 litre atm K^ 1 mol^ 1 ...

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Standard XII

Chemistry

Derivation of Kp and Kc

At 700 K, the...

Question

At 700 K, the equilibrium constant $K_{p},$ for the reaction
$2 S O_{3} (g) ⇌ 2 S O_{2} (g) + O_{2} (g)$
is $1.8 \times 10^{- 3} k P a .$ What is the numerical value of $K_{C}$ for this reaction at the same temperature?
$R = 0.0821 L a t m m o l^{- 1} K^{- 1}$

3.09 \times 10^{- 7} m o l e l i t r e^{- 1}

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9.03 \times 10^{- 7} m o l e l i t r e^{- 1}

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5.05 \times 10^{- 9} m o l e l i t r e^{- 1}

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5.05 \times 10^{- 5} m o l e l i t r e^{- 1}

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Solution

The correct option is A $3.09 \times 10^{- 7} m o l e l i t r e^{- 1}$
We know the relationship
$K_{P} = K_{C} (R T)^{△ n}$
Here $K_{P} = 1.80 \times 10^{- 3}$
$K_{P} = \frac{1.8 \times 10^{- 3}}{101.3} a t m$
$= 1.78 \times 10^{- 5} a t m$
$R = 0.0821$ litre atm $K^{- 1} m o l^{- 1}$
$△ n = 3 - 2 = 1$
$T = 700 K$
$K_{C} = \frac{K_{P}}{(R T)^{△ n}}$
putting the values,
$K_{C} = \frac{1.78 \times 10^{- 5}}{0.0821 \times 700}$
$= 3.09 \times 10^{- 7} m o l e l i t r e^{- 1} .$

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Similar questions

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. At 700 K, the equilibrium constant Kp, for the reaction

2 S O_{3} (g) ⇌ 2 S O_{2} (g) + O_{2} I s 1.8 \times 10^{- 3} K P a

. What is the numerical value in moles per liter of Kc for the reaction at the same temperature?

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)

1.80 \times 10^{- 3} k p a .

What is the numeriacal value in moles per litre of

K_{c}

for this reaction at the same temperaure?

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. At 700 K, the equilibrium constant,

K_{p}

, for the reaction

2 S O_{3} (g) ⇌ 2 S O_{2} (g) + O_{2} (g) i s 1.8 \times 10^{- 3}

atm. The value of

K_{c}

for the above reaction at the same temperature in moles per litre would be:

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At 700 K, the equilibrium constant Kp, for the reaction 2SO3(g)⇌2SO2(g)+O2(g) is 1.8×10−3kPa. What is the numerical value of KC for this reaction at the same temperature? R=0.0821 L atm mol−1 K−1

The correct option is A 3.09×10−7 mole litre−1We know the relationship KP=KC(RT)△n Here KP=1.80×10−3 KP=1.8×10−3101.3atm =1.78×10−5 atm R=0.0821 litre atm K−1 mol−1 △n=3−2=1 T=700 K KC=KP(RT)△n putting the values, KC=1.78×10−50.0821×700 =3.09×10−7 mole litre−1.

At 700 K, the equilibrium constant $K_{p},$ for the reaction
$2 S O_{3} (g) ⇌ 2 S O_{2} (g) + O_{2} (g)$
is $1.8 \times 10^{- 3} k P a .$ What is the numerical value of $K_{C}$ for this reaction at the same temperature?
$R = 0.0821 L a t m m o l^{- 1} K^{- 1}$