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sladkih [1.3K]
2 years ago
10

Methyl iodide reacts irreversibly with azide ion with rate = k[CH3I][N3–]. CH3I(aq) + N3–(aq) → CH3N3(aq) + I–(aq) The reaction

is carried out with an initial concentration of CH3I of 0.01 M. Which statement about the reaction is correct?
Chemistry
1 answer:
sergij07 [2.7K]2 years ago
4 0

Answer:

(D) The reaction cannot take place in a single elementary step

Explanation:

<em>Statements are:</em>

<em>(A) The time it takes for [CH3I] to decrease to 0.005 M is independent of [N3-], as long as [N3] >> [CH3I]. </em>

<em>B) If the initial concentrations of azide and CH3I are equal, then it takes half as long for [CH3I] to decrease to 0.005 M as it does for it to decrease from 0.005 M to 0.0025 M. </em>

<em>(C) The reaction rate is significantly smaller if excess I- is added to the solution. </em>

<em>(D) The reaction cannot take place in a single elementary step.</em>

<em />

The rate of the reaction is:

rate = k[CH3I][N3–].

That means rate depends of concentration of CH₃I as much as N₃⁻ concentration

(A) The time it takes for [CH3I] to decrease to 0.005 M is independent of [N3-], as long as [N3] >> [CH3I]. FALSE. The reaction rate depends of N₃⁻ as much as CH₃I

B) If the initial concentrations of azide and CH3I are equal, then it takes half as long for [CH3I] to decrease to 0.005 M as it does for it to decrease from 0.005 M to 0.0025 M. FALSE. Reaction is second-order. Half-life is 1/K[A]₀. If initial concentration is 0.1M, to a concentration of 0.005M it takes:

1/K*0.1. If initial concentration is 0.005M it takes 1/K*0.005. That means it takes half to decrease from 0.005M to 0.0025 as it does for it to decrease from 0.01M to 0.005M.

(C) The reaction rate is significantly smaller if excess I- is added to the solution. FALSE. Reaction rate is independent of I⁻

(D) The reaction cannot take place in a single elementary step. TRUE. As this reaction is a single-replacement reaction implies the formation  of 1 C-N bond. But also the rupture of the C-I bond is impossible to explain this kind of reaction in a single elementary step.

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Phosphorous acid, H3PO3(aq), is a diprotic oxyacid that is an important compound in industry and agriculture. K pKa1 K pKa2 1.30
FrozenT [24]

Answer:

* Before addition of any KOH:

pH = 0,0301

*After addition of 25.0 mL KOH:

pH = 1,30

*After addition of 50.0 mL KOH:

pH = 2,87

*After addition of 75.0 mL KOH:

pH = 6,70

*After addition of 100.0 mL KOH:

pH = 10,7

Explanation:

H₃PO₃ has the following equilibriums:

H₃PO₃ ⇄ H₂PO₃⁻ H⁺

k = [H₂PO₃⁻] [H⁺] / [H₃PO₃] k = 10^-(1,30) <em>(1)</em>

H₂PO₃⁻ ⇄ HPO₃²⁻ + H⁺

k = [HPO₃²⁻] [H⁺] / [H₂PO₃⁻] k = 10^-(6,70) <em>(2)</em>

Moles of H₃PO₃ are:

0,0500L×(1,8mol/L) = 0,09 moles of H₃PO₃

* Before addition of any KOH:

Using (1), moles in equilibrium are:

H₃PO₃: 0,09-x

H₂PO₃⁻: x

H⁺: x

Replacing:

10^{-1.30} = \frac{x^2}{0.09-x}

4.51x10⁻³ - 0.050x -x² = 0

The right solution of x is:

x = 0.0466589

As volume is 0,050L

[H⁺] = 0.0466589moles / 0,050L = 0,933M

As pH = -log [H⁺]

<em>pH = 0,0301</em>

*After addition of 25.0 mL KOH:

0,025L×1,8M = 0,045 moles of KOH that reacts with H₃PO₃ thus:

KOH + H₃PO₃ → H₂PO₃⁻ + H₂O

That means moles of KOH will be the same of H₂PO₃⁻ and moles of H₃PO₃ are 0,09moles - 0,045moles = 0,045moles

Henderson-Hasselbalch formula is:

pH = pka + log₁₀ [A⁻] /[HA]

Where A⁻ is H₂PO₃⁻ and HA is H₃PO₃.

Replacing:

pH = 1,30 + log₁₀ [0,045mol] / [0,045mol]

<em>pH = 1,30</em>

*After addition of 50.0 mL KOH:

The addition of 50.0 mL KOH consume all H₃PO₃. Thus, in the solution you will have just H₂PO₃⁻. Thus, moles in solution for the equilibrium will be:

H₂PO₃⁻: 0,09-x

HPO₃²⁻: x

H⁺: x

Replacing:

10^{-6.70} = \frac{x^2}{0.09-x}

1.8x10⁻⁸ - 2x10⁻⁷x - x² = 0

The right solution of x is:

x = 0.000134064

As volume is 50,0mL + 50,0mL = 100,0mL

[H⁺] = 0.000134064moles / 0,100L = 1.34x10⁻³M

As pH = -log [H⁺]

<em>pH = 2,87</em>

*After addition of 75.0 mL KOH:

Applying Henderson-Hasselbalch formula you will have 0,045 moles of both H₂PO₃⁻ HPO₃²⁻ and pka: 6,70:

pH = 6,70 + log₁₀ [0,045mol] / [0,045mol]

<em>pH = 6,70</em>

*After addition of 100.0 mL KOH:

You will have just 0,09moles of HPO₃²⁻, the equilibrium will be:

HPO₃²⁻ + H₂O ⇄ H₂PO₃⁻ + OH⁻ with kb = kw/ka = 1x10⁻¹⁴/10^-(6,70) = 5,01x10⁻⁸

kb = [H₂PO₃⁻] [OH⁻] / [HPO₃²⁻]

Moles are:

H₂PO₃⁻: x

OH⁻: x

HPO₃²⁻: 0,09-x

Replacing:

5.01x10^{-8} = \frac{x^2}{0.09-x}

4.5x10⁻⁹ - 5.01x10⁻⁸x - x² = 0

The right solution of x is:

x = 0.000067057

As volume is 50,0mL + 100,0mL = 150,0mL

[OH⁻] = 0.000067057moles / 0,150L = 4.47x10⁻⁴M

As pH = 14-pOH; pOH = -log [OH⁻]

<em>pH = 10,7</em>

<em></em>

I hope it helps!

6 0
2 years ago
Two protons and two neutrons are released as a result of this reaction.
Murljashka [212]

<u>Answer:</u> The particle released in the given reaction is one alpha particle

<u>Explanation:</u>

In a nuclear reaction, the total mass and total atomic number remains the same.

For the given fission reaction:

^{222}_{86}\textrm{Rn}\rightarrow ^A_Z\textrm{X}+^{218}_{84}\textrm{Po}

  • <u>To calculate A:</u>

Total mass on reactant side = total mass on product side

222 = A + 218

A = 4

  • <u>To calculate Z:</u>

Total atomic number on reactant side = total atomic number on product side

86 = Z + 84

Z = 2

The isotopic symbol of unknown element is _{2}^{4}\textrm{He}. Another name of helium atom is alpha particle.

Hence, the particle released in the given reaction is one alpha particle

3 0
2 years ago
Read 2 more answers
A pan containing 40 grams of water was allowed to cool from a temperature of 91.0C. If the amount of heat repressed is 1,300 jou
Vinvika [58]

Answer:

83°C

Explanation:

The following were obtained from the question:

M = 40g

C = 4.2J/g°C

T1 = 91°C

T2 =?

Q = 1300J

Q = MCΔT

ΔT = Q/CM

ΔT = 1300/(4.2x40)

ΔT = 8°C

But ΔT = T1 — T2 (since the reaction involves cooling)

ΔT = T1 — T2

8 = 91 — T2

Collect like terms

8 — 91 = —T2

— 83 = —T2

Multiply through by —1

T2 = 83°C

The final temperature is 83°C

3 0
2 years ago
what mass of lithium chloride, licl must be dissolved to make a 0.194M solution that has volume of 1.00 l
Bogdan [553]
MXV= (0.194M)(1.00L)=0.194moles
42.39LiClg/molex0.194moles=8.2g LiCl
6 0
2 years ago
When 10 g of diethyl ether is converted to vapor at its boiling point, about how much heat is absorbed? (c4h10o, δhvap = 15.7 kj
vagabundo [1.1K]

Answer:

ΔH= 3KJ

Explanation:

The total heat absorbed is the total energy in the process, and that is in form of entalpy.

ΔH = q + ΔHvap, where q is the heat necessary for elevate the temperature of dietil ether. Suppose the initial temperature is room temperature (25ºC=298 K), then

q= 10g x2.261 J/gK x(310 K - 298K)= 271.32 J= 0.3 kJ

Then

ΔHvap = 10g C4H10O x (1 mol C4H10O/74.12 g C4H10O) x( 15.7 KJ/ 1 mol C4H10O) = 2.12 KJ

ΔH= 2.5KJ ≈ 3KJ

5 0
2 years ago
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