Equilibrium

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New answer posted•a year ago

Equilibrium Class 11th

For the reaction A + B ? 2C
The value of equilibrium constant is 100 at 298K. If the initial concentration of all the three species is 1M each, then the equilibrium concentration of C is x × 10?¹M. The value of x is ______ (Nearest integer).

4 Views|Posted a year ago

Asked by Raj Pandey

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Answered by

Raj Pandey | Contributor-Level 9

a year ago

A + B? 2C
Initial: 1, 1
At eq: 1-x, 1+2x
K = [C]²/ ( [A] [B]) = (1+2x)²/ (1-x)² = 100
(1+2x)/ (1-x) = 10
1+2x = 10-10x => 12x = 9 => x = 3/4
[C] = 1+2x = 1+2 (3/4) = 1+1.5 = 2.5M = 25 × 10? ¹M

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New answer posted•a year ago

Equilibrium Class 11th

PCl₅ ⇌ PCl₃ + Cl₂ Kc = 1.844. 3.0 moles of PCl₅ is introduced in a 1L closed reaction vessel at 380K. The number of moles of PCl₅ at equilibrium is _______ × 10⁻³. (Round off to the Nearest integer)

23 Views|Posted a year ago

Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

PCl? (g)? PCl? (g) + Cl? (g); Kc = 1.844
t=0: 3, 0, 0
equilibrium: 3-x, x
Kc = x² / (3-x) = 1.844
x² + 1.844x - 5.532 = 0
x = (-1.844 + √ (1.844² - 4 (1) (-5.532)/2 = (-1.844 + √25.528)/2 ≈ 1.604
At equilibrium number of moles of PCl? = (3 - 1.604) = 1.396 mol
= 1396 × 10? ³ mol

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New answer posted•a year ago

Equilibrium Class 11th

At 1990K and 1 atm pressure, there are equal number of Cl₂ molecules and Cl atoms in the reaction mixture. The value of K_P for the reaction $?$ Cl_2(g)  2Cl_(g) under the above conditions is x × 10^-1. The value of x is______ (Rounded off to the nearest integer)

7 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Initially -> 1 mol -

At eq. &nb

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New answer posted•a year ago

Equilibrium Class 11th

The pH of ammonium phosphate solution, if pk_a of phosphoric acid and pK_b of ammonium hydroxide are 5.23 and 4.75 respectively, is…………….

4 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

pH = $\frac{1}{2} [p K_{w} + p K_{a} - p K_{b}]$ $\frac{}{}_{}_{}_{}$

$= \frac{1}{2} [14 + 4.75 - 5.23]$

$= 6.76 ≅ 7$

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New answer posted•a year ago

Equilibrium Class 11th

For the reaction A_(g) ® B_(g) the value of the equilibrium constant at 300K and 1 atm is equal to 100.0 The value of for the reaction at 300K and 1 atm in J mol^-¹ is -xR, where x is_________.

(Rounded off to the nearest integer)

[R = 8.31 J mol^-¹ K^-¹ and In 10 = 2.3]

2 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

A (g) -> B (g) K_P = 100

$Δ G$ at 300 K and 1 atm

Using

$Δ G = - R T l n K_{P}$

$Δ G = - R \times 300 l n 100$

=-R × 300 × 2 × 2.3

$Δ G = - 1380 R$

So; x = 1380.

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New answer posted•a year ago

Equilibrium Class 11th

The solubility of AgCN in a buffer solution of pH = 3 is x. The value of x is:

[Assume : No cyano complex is fomed ; k_sp (AgCN) = 2.2 × 10^-16 and K_a (HCN) = 6.2 × 10^-10]

6 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$\underset{S S}{A g C N_{(S)} ? A {g^{+}}_{(a q)} + C {N^{-}}_{(a q)}}$

$C N^{-} + H^{+} ? H C N$

Before reaction S 10^-3 0

After reaction 0 &

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New answer posted•a year ago

Equilibrium Class 11th

The solubility of Ca(OH)₂ in water is:

[Given : The solubility product of Ca(OH)₂ in water = 5.5 × 10^-6]

5 Views|Posted a year ago

Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

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New answer posted•a year ago

Equilibrium Class 11th

For the reaction $2 A ? B + C, K_{C} = 4 \times 1 0^{- 3}$ . At a given time, the composition of reaction mixture is: $[A] = [B] = [C] = 2 \times 1 0^{- 3} M$ . Then, which of the following is correct?

5 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

$2 ? + C, K_{C} = 4 \times 1 0^{- 3}$

At a given time $t, Q_{c}$ is to be calculated and been compared with $K_{c}$ .

$Q_{C} = \frac{[B] [C]}{{[A]}^{2}} = \frac{(2 \times 1 0^{- 3}) (2 \times 1 0^{- 3})}{{(2 \times 1 0^{- 3})}^{2}}$

$Q_{C} = 1$

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New answer posted•a year ago

Equilibrium Class 11th

20 mL of 0.1 M NH₄Oh is mixed with 40 mL of 0.05 M HCl. The pH of the mixture is nearest to: (Given K_b (NH₄OH) = 1 × 10^-5, log2 = 0.30, log3 = 0.48, log5 = 0.69, log7 = 0.84, log11 = 1.04

6 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$[N H_{4} C l] = \frac{2}{60} = \frac{1}{30} M$

$p H = 7 - \frac{1}{2} P K_{b} - \frac{1}{2} l o g C$

$= 7 - \frac{5}{2} - \frac{1}{2} l o g (\frac{1}{30}) = 5.24$

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New answer posted•a year ago

Equilibrium Class 11th

Ka for butyric acid (C₃H₇COOH) is 2 × 10^-⁵. The pH of 0.2 M solution of butyric acid is ____________× 10^-¹ (Nearest integer)

[Given log2 = 0.30]

72 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Ka for C₃H₇COOH = 2 × 10^-⁵

$p K a = - l o g (2 \times 1 0^{- 5}) = 5 - l o g 2$

=5 – 0.3 = 4.7

pH of 0.2 (M) solution =

$p H = \frac{p K a - l o g C}{2}$

$= \frac{1}{2} (4.7) - \frac{1}{2} l o g (0.2)$

$p H = 27 \times 1 0^{- 1}$

Ans 27

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