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DedPeter [7]
3 years ago
5

The first-order rate constant for the reaction of methyl chloride (CH3Cl) with water to produce methanol (CH3OH) and hydrochlori

c acid (HCl) is 3.32 × 10−10 s−1 at 25°C. Calculate the rate constant at 48.5°C if the activation energy is 116 kJ/mol.
Chemistry
1 answer:
Dvinal [7]3 years ago
7 0

Answer:

K(48.5°C) = 1.017 E-8 s-1

Explanation:

  • CH3Cl + H2O → CH3OH + HCl

at T1 = 25°C (298 K) ⇒ K1 = 3.32 E-10 s-1

at T2 = 48.5°C (321.5 K) ⇒ K2 = ?

Arrhenius eq:

  • K(T) = A e∧(-Ea/RT)
  • Ln K = Ln(A) - [(Ea/R)(1/T)]

∴ A: frecuency factor

∴ R = 8.314 E-3 KJ/K.mol

⇒ Ln K1 = Ln(A) - [Ea/R)*(1/T1)]..........(1)

⇒ Ln K2 = Ln(A) - [(Ea/R)*(1/T2)].............(2)

(1)/(2):

⇒ Ln (K1/K2) = (Ea/R)* (1/T2-1/T1)

⇒ Ln (K1/K2) = (116 KJ/mol/8.3134 E-3 KJ/K.mol)*(1/321.5 K - 1/298 K)

⇒ Ln (K1/K2) = (13952.37 K)*(- 2.453 E-4 K-1)

⇒ Ln (K1/K2) = - 3.422

⇒ K1/K2 = e∧(-3.422)

⇒ (3.32 E-10 s-1)/K2 = 0.0326

⇒ K2 = (3.32 E-10 s-1)/0.0326

⇒ K2 = 1.017 E-8 s-1

You might be interested in
if it takes 54 mL of 0.1 NaOH to neutralize 125 mL of an HCL solution, what is the concentration of HCL?
OleMash [197]

Answer:

0.0432M

Explanation:

We begin by writing a balanced equation for the reaction. This is illustrated below:

NaOH + HCl —> NaCl + H2O

From the equation above,

The number of mole of the acid (nA) = 1

The number of mole of the base (nB) = 1

Data obtained from the question include:

Vb (volume of the base) = 54mL

Cb (concentration of the base) = 0.1M

Va (volume of the acid) = 125mL

Ca ( concentration of the acid) =?

Using CaVa/CbVb = nA/nB, the concentration of the acid can easily be obtained as shown below:

CaVa/CbVb = nA/nB

Ca x 125 / 0.1 x 54 = 1

Cross multiply to express in linear form:

Ca x 125 = 0.1 x 54

Divide both side by 125

Ca = (0.1 x 54) / 125

Ca = 0.0432M

Therefore, the concentration of the acid is 0.0432M

3 0
3 years ago
how many grams of antifreeze would be required per 500 g of water to prevent the water from feezing at a temperature of -39° C​
andrezito [222]

Answer:

333.7g of antifreeze

Explanation:

Freezing point depression in a solvent (In this case, water) occurs by the addition of a solute. The law is:

ΔT = Kf × m × i

Where:

ΔT is change in temperature (0°C - -20°C = 20°C)

Kf is freezing point depression constant (1.86°C / m)

m is molality of solution (moles solute / 0.5 kg solvent -500g water-)

i is Van't Hoff factor (1, assuming antifreeze is ethylene glycol -C₂H₄(OH)₂)

Replacing:

20°C = 1.86°C / m  × moles solute / 0.5 kg solvent × 1

5.376 = moles solute

As molar mass of ethylene glycol is 62.07g/mol:

5.376 moles × (62.07g / 1mol) = <em>333.7g of antifreeze</em>.

4 0
3 years ago
If the reactants contain one chlorine molecule, do you know how many chlorine atoms will be in the product? Include conversion o
max2010maxim [7]

Answer:

The number of Chlorine atoms in the product is 2.

Explanation:

The law of conservation of mass states that matter can neither be created nor destroyed in a chemical reaction.

The reactants contain one chlorine molecule(Cl_{2}) which has two chlorine atoms.

Then, according to law of conservation of matter, the product must contain two chlorine atoms.

4 0
3 years ago
The average score for games played in the NFL is 21.1 and the standard deviation is 8.9 points. 46 games are randomly selected.
kaheart [24]

Answer:

A) P(21.4317 < ¯x < 22.7561) = 0.2975

B) Q1 for the ¯x distribution = 21.9844

Explanation:

The Central Limit theorem allows us to say that

Sample mean = Population mean = 21.1 points

Mean of sampling distribution = σₓ = (σ/√n)

σ = population standard deviation = 8.9 points

n = sample size = 46

σₓ = (8.9/√46) = 1.3122334098 = 1.3122

A) P(21.4317 < ¯x < 22.7561) =

This is a normal distribution problem

To find this probability, we will use the normal probability tables

We first normalize/standardize 21.4317 and 22.7561.

The standardized score of any value is that value minus the mean divided by the standard deviation.

For 21.4317

z = (x - μ)/σ = (21.4317 - 21.1)/1.3122 = 0.25

For 22.7561

z = (x - μ)/σ = (22.7561 - 21.1)/1.3122 = 1.26

The required probability

P(21.4317 < ¯x < 22.7561) = P(0.25 < z < 1.26)

Checking the tables

P(21.4317 < ¯x < 22.7561) = P(0.25 < z < 1.26)

= P(z < 1.26) - P(z < 0.25)

= 0.89617 - 0.59871

= 0.29746 = 0.2975 to 4 d.p.

B) Q1 for the distribution is the first quartile. The first quartile is greater than 25% of the distribution.

P(x > Q1) = 0.25

Let the z-score that corresponds to Q1 be z'

P(x > Q1) = P(z > z') = 0.25

But P(z > z') = 1 - P(z ≤ z') = 0.25

P(z ≤ z') = 1 - 0.25 = 0.75

From the normal distribution tables,

z' = 0.674

z' = (Q1 - μ)/σ

0.674 = (Q1 - 21.1)/1.3122

Q1 = 0.674×1.3122 + 21.1 = 21.9844228 = 21.9844 to 4 d.p.

Hope this Helps!!!

7 0
3 years ago
PLZ HELPPPP TIMED TESTT!!!!!!?????
Leya [2.2K]

Answer:

my probably best answer is B

5 0
3 years ago
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