Equilibrium

pH for 0.01 M NaOH solution is

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0.01 M NaOH,

M = 1 * 10^-2

pOH = 2

pH = 2

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K_p for the given reaction is (36 * 10^–2atm^–1) Find out K_c (M^–1) (nearest integer).

(2NO₂ N₂O₄)

(R=0.0821 atm. L/mol. K)

(T = 300K)

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K_p = K_c (RT)^D^ng

36 * 10^–2 = K_c (0.0821 * 300)^–1

K_c = 0.36 * 0.0821 * 300 = 8.86 » 9

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A(g) ->B(g) + $\frac{1}{2}$ C(g)

In the about reaction, the correct relation between K_p, a and equilibrium pressure (p) is

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Contributor-Level 10

A(g) ->B(g) + $\frac{1}{2}$ (g)

Initial moles n 0 0

Eqb. moles n(1 – a) $\frac{n α}{2}$ n_a

total moles = $n (1 + \frac{α}{2})$

Eqb. pressure $\frac{(1 - α) p}{1 + \frac{α}{2}}$ $\frac{α p}{1 + \frac{α}{2}}$ $\frac{(\frac{α}{2}) p}{1 + \frac{α}{2}}$

$\frac{K_{p} = \frac{α p}{(1 + \frac{α}{2})} * {[\frac{α p}{(2 + α)}]}^{\frac{1}{2}}}{\frac{(1 + α) p}{1 + \frac{α}{2}}}$

$K_{p} = \frac{α^{\frac{3}{2}} p^{\frac{1}{2}}}{{(2 + α)}^{\frac{1}{2}} (1 - α)}$ &nbs

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A(g) ->B(g) + $\frac{1}{2}$ (g)

Initial moles n 0 0

Eqb. moles n(1 – a) $\frac{n α}{2}$ n_a

total moles = $n (1 + \frac{α}{2})$

Eqb. pressure $\frac{(1 - α) p}{1 + \frac{α}{2}}$ $\frac{α p}{1 + \frac{α}{2}}$ $\frac{(\frac{α}{2}) p}{1 + \frac{α}{2}}$

$\frac{K_{p} = \frac{α p}{(1 + \frac{α}{2})} * {[\frac{α p}{(2 + α)}]}^{\frac{1}{2}}}{\frac{(1 + α) p}{1 + \frac{α}{2}}}$

$K_{p} = \frac{α^{\frac{3}{2}} p^{\frac{1}{2}}}{{(2 + α)}^{\frac{1}{2}} (1 - α)}$

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Which of the following equilibrium will shift in forward direction on applying pressure?

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On increasing pressure, equilibrium moves in that direction where number of gaseous moles decreases.

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The ionization constant of HCN is 4.5 * 10^–10 at 298 K. Hydrolysis constant of sodium cyanide is

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The pH of 1 M CH₃COONa solution in water will be nearly (pKa for CH₃COOH = 4.74)

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Equilibrium Class 11

What is the weightage of Equilibrium Class 12 Chemistry in NEET and JEE Main exam?

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In NEET exam, the weightage of Equilibrium is around 3% to 6% and in JEE Main exam, the weightage is 4% to 6.6%.

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2 weeks ago

Equilibrium Class 11

What are the conditions required for equilibrium.

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There are two conditions needed for equilibrium. The net force and net torque acting upon an object must be zero.

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2 weeks ago

Equilibrium NEET

Which are the types of equilibrium constants?

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There are four types of equilibrium constants: Formation constants, stability constants, association constants, and binding constants.

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2 weeks ago

Equilibrium NEET

Is class 12 chemistry equilibrium an easy chapter?

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