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Chemical Equilibrium

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2 months ago

Chemical 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]

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Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

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2 months ago

Chemical Equilibrium Equilibrium

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…………….

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alok kumar singh

Contributor-Level 10

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

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

$= 6.76 ≅ 7$

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2 months ago

Chemical Equilibrium Equilibrium

What is dynamic equilibrium?

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Swayam Gupta

Contributor-Level 6

It happens in reversible reactions when the rate of the forward reaction becomes equal to the rate of the backward reaction. Result in the same concentration of reactants and product.

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2 months ago

Chemical Equilibrium Equilibrium

What is Le Chatelier's principle NCERT?

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Swayam Gupta

Contributor-Level 6

If the conditions of equilibrium are changed, it shifts to oppose the change. For example, in Haber's process, high pressure favors NH? formation.

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2 months ago

Chemical Equilibrium Equilibrium

What is the equilibrium constant?

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Swayam Gupta

Contributor-Level 6

The equilibrium constant is the ratio of the concentrations of products to reactants, each raised to the power of their stoichiometric coefficients. For reversible reactions.

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2 months ago

Chemical Equilibrium Equilibrium

4.0 moles of argon and 5.0 moles of PCl₅ are introduced into an evacuated flask of 100 litre capacity at 610 K. The system is allowed to equilibrate. At equilibrium, the total pressure of mixture was found to be 6.0 atm. The K_p for the reaction is

[Given R = 0.082 L atm K^-1 mol^-1]

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Vishal Baghel

Contributor-Level 10

Moles of PCl₅ = 5 mol

Moles of Ar = 4 mol

Total no of moles = 9 moles

$P_{T o t a l} = \frac{n R T}{V} = \frac{9 * 0.0821 * 610}{100} = 4.5 a t m$

$P_{P C l_{5}} = X_{P C l_{5}} * P_{T} = \frac{5}{9} * 4.5 = 2.5 a t m$

$P_{A r} = X_{A r} * P_{T} = \frac{4}{9} * 4.5 = 2 a t m$

$P C l_{5} \overset{}{?} P C l_{3} + C l_{2}$

2.5 0 0

2.5 - P P P

$P_{T o t a l} = 2.5 - P + P + P + P_{A r} = 6$

P = 1.5

...more

Moles of PCl₅ = 5 mol

Moles of Ar = 4 mol

Total no of moles = 9 moles

$P_{T o t a l} = \frac{n R T}{V} = \frac{9 * 0.0821 * 610}{100} = 4.5 a t m$

$P_{P C l_{5}} = X_{P C l_{5}} * P_{T} = \frac{5}{9} * 4.5 = 2.5 a t m$

$P_{A r} = X_{A r} * P_{T} = \frac{4}{9} * 4.5 = 2 a t m$

$P C l_{5} \overset{}{?} P C l_{3} + C l_{2}$

2.5 0 0

2.5 - P P P

$P_{T o t a l} = 2.5 - P + P + P + P_{A r} = 6$

P = 1.5 atm

$K_{P} = \frac{1.5 * 1.5}{1} = 2.25$

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New answer posted

2 months ago

Chemical Equilibrium Equilibrium

[Given R = 0.082 L atm K^-¹ mol^-¹]

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alok kumar singh

Contributor-Level 10

Moles of PCl₅ = 5 mol

Moles of Ar = 4 mol

Total no of moles = 9 moles

$P_{T o t a l} = \frac{n R T}{V} = \frac{9 * 0.0821 * 610}{100} = 4.5 a t m$

$P_{P C l_{5}} = X_{P C l_{5}} * P_{T} = \frac{5}{9} * 4.5 = 2.5 a t m$

$P_{A r} = X_{A r} * P_{T} = \frac{4}{9} * 4.5 = 2 a t m$

$P C l_{5} \overset{}{?} P C l_{3} + C l_{2}$

2.5 0 0

2.5 - P P P

$P_{T o t a l} = 2.5 - P + P + P + P_{A r} = 6$

P = 1.5

...more

Moles of PCl₅ = 5 mol

Moles of Ar = 4 mol

Total no of moles = 9 moles

$P_{T o t a l} = \frac{n R T}{V} = \frac{9 * 0.0821 * 610}{100} = 4.5 a t m$

$P_{P C l_{5}} = X_{P C l_{5}} * P_{T} = \frac{5}{9} * 4.5 = 2.5 a t m$

$P_{A r} = X_{A r} * P_{T} = \frac{4}{9} * 4.5 = 2 a t m$

$P C l_{5} \overset{}{?} P C l_{3} + C l_{2}$

2.5 0 0

2.5 - P P P

$P_{T o t a l} = 2.5 - P + P + P + P_{A r} = 6$

P = 1.5 atm

$K_{P} = \frac{1.5 * 1.5}{1} = 2.25$

less

0 0

New answer posted

2 months ago

Chemical 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

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Raj Pandey

Contributor-Level 9

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

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

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

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New answer posted

2 months ago

Chemical Equilibrium Equilibrium

An equilibrium mixture contains 2.0 moles of A(g) and 4.0 moles of B(g). Now, 2.0 moles of A(g) is added in this equilibrium mixture and the system is allowed to re-achieve equilibrium at constant volume and temperature. Now, the volume of system is doubled at constant temperature. What should be the moles of B(g) at new equilibrium? The reaction involved is

$A (g) \overset{}{?} 2 B (g)$ .

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Vishal Baghel

Contributor-Level 10

$A (g) ? 2 B (g) .$

$E_{m} \frac{2 m o l}{V l} \frac{4 m o l}{V l}$

$N e w {E_{q}}^{m} \frac{4 - x}{V} \frac{4 + 2 x}{V}$

$V * 2 \frac{4 - x}{2 V} \frac{4 - 2 x}{2 V}$

New Eqm $\frac{4 - x - y}{2 V} \frac{4 + 2 x + 2 y}{2 V}$

= $\frac{4 - Z}{2 V} = \frac{4 + 2 Z}{2 V} (Z = x + y)$

Now, KC = $\frac{^{}}{}$ $\frac{{[B]}^{2}}{[A]} =$ constant

$o r, \frac{{(\frac{4}{V})}^{2}}{(\frac{2}{V})} = \frac{(\frac{4 + 2 Z}{2 V})}{(\frac{4 - Z}{2 V})} \Rightarrow Z = 1.29$

mole of B at new Eqm = 4 + 2Z = 6.58

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New answer posted

2 months ago

Chemical Equilibrium Class 11th

Identify the numbers of amphiprotic species in the following list?
H?O?, HPO?²?, HCO?, H?O, HPO?²?, H?PO?, H?PO?, H?PO?, PO?³?, NH?

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Raj Pandey

Contributor-Level 9

Amphiprotic species are those which behave like acid and base both. They can donate and accept a proton.

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