Chemical Equilibrium

The volume (in mL) of 0.01M NaOH required to neutralise 20ml of 0.01M ortho-phosphoric acid is ____.

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Contributor-Level 9

$N_{1} V_{1} = N_{2} V_{2}$

$0.01 * V_{1} = 0.01 * 20 * 3$

$V_{1} = 60 m L$

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2 months ago

A container contains O?, He and SO? gases such that mass of SO? is double the mass of O? and mass of He is half the mass of O? gas. The sum of mole fraction of all the gases is:

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Contributor-Level 9

Sum of mole fraction of all the gases is 1 .

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2 months ago

For the reaction NH₄Cl(g) ⇌ NH₃(g) + HCl(g)

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Contributor-Level 10

From Reaction

Δn? = 2 – 1 = 1

Kp = Kc (RT)^Δn?

Kp = Kc (RT)¹

Kc = Kp (RT)? ¹

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2 months ago

Arrange the following solutions in the decreasing order of pH :
(1) 0.1 NaOH
(2) 0.1 M NaCl
(3) 0.1 M H₂SO₄
(4) 0.1 M NH₄Cl

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Contributor-Level 10

All have same molarity, pH ∝ 1/ (Acidic strength) ∝ Basic strength

H? SO? → Acidic

NH? Cl (Salt of weak base and strong acid)

Acidic but less than H? SO?

NaCl → (Neutral solution)

NaOH → Basic (Strong Base)

So NaOH > NaCl > NH? Cl > H? SO?

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2 months ago

The increasing order of pk_b values of the following compounds is:

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Contributor-Level 10

and basic strength ∝ 1/pK? ∝ K? ∝ +R ∝ 1/ (-R) ∝ +I ∝ 1/ (-I)

So basic strength order ⇒ IV > III > II > I, so pK? order is IV < III < II < I

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2 months ago

The standard free energy change $27 ° C$ for 50% dissociation of N₂O₄ into NO₂ at $(Δ G °)$ and 1 atm pressure is -xJ mol^-¹. The value of x is_________. (Nearest Integer)

[Given: R= 8.31 J k^-¹ mol^-¹, log 1.33 = 0.1239, In 10 = 2.3]

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Contributor-Level 9

$N_{2} O_{4} ? 2 N O_{2}$

$\underset{= 0.5 m o l}{(1 - 0.5)}$ $2 * 0.5 = 1 m o l$

$k_{P} = \frac{{(\frac{1}{1.5} * 1)}^{2}}{(\frac{0.5}{1.5} * 1)}$

$= \frac{4}{3} = 1.33$

$= - 710.15 J / m o l$

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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)

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alok kumar singh

Contributor-Level 10

$C l_{2 (g)} ? 2 C l_{(g)}$

Initially -> 1 mol -

At eq. 1-x mol 2x mol

Here; molecules of Cl₂ = atoms of Cl

i.e moles of Cl₂ = moles of Cl

So : 1 – x = 2x x = 1/3

Moles of Cl₂ at equ

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$C l_{2 (g)} ? 2 C l_{(g)}$

Initially -> 1 mol -

At eq. 1-x mol 2x mol

Here; molecules of Cl₂ = atoms of Cl

i.e moles of Cl₂ = moles of Cl

So : 1 – x = 2x x = 1/3

Moles of Cl₂ at equibrium = $\frac{2}{3}$

Moles of Cl at equilibrium = $\frac{2}{3}$

Total moles = $\frac{4}{3}$

Now: $P_{C l_{2}} = \frac{2}{4} * 1 a t m$

$= \frac{1}{2} a t m$

$P_{C l} = \frac{2}{4} * 1 a t m = \frac{1}{2} a t m$

$K_{P} = \frac{{(P_{C l})}^{2}}{P_{C l_{2}}} = \frac{{(\frac{1}{2})}^{2}}{\frac{1}{2}} = \frac{1}{2}$

$K_{P} = 0.5 = 5 * 1 0^{- 1}$

Here : x = 5

less

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2 months ago

0.25 M solution of pyridinium chloride C₅H₆N⁺Cl⁻ was found to have a pH of 2.699. if the value of Kₑ of pyridine C₅H₅N is a * 10⁻¹⁰, then the value of a is (given, 10⁻².⁶⁹⁹ = 2 * 10⁻³)

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Contributor-Level 10

BCl → B? + Cl?
B? + H? O? BOH + H? , K? = K? /K?
0.25 – –
(0.25 – x) x
Given, pH = 2.7 = [H? ] = 2 * 10? ³
∴ x²/ (0.25) = (10? ¹? )/K?
= 4 * 10? * 4 * K? = 10? ¹?
= K? = (1/16) * 10? = 6.25 * 10? ¹?

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2 months ago

40% of HI undergoes decomposition to H2 and I2 at 300K. $^{}$ for this decomposition reaction at one atmosphere pressure is_____________J mol1 (nearest integer)

(Use R = 8.31 JK1 mol1; log 2= 0.3010, In 10 = 2.3, log 3 = 0.477)

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