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2 years ago

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Given that the initial rate constant is 0.0110s−1 at an initial temperature of 21 ∘C , what would the rate constant be at a temp

erature of 150∘C for the same reaction described in Part A?
This was part A
The activation energy of a certain reaction is 45.5 kJ/mol . At 21 ∘C , the rate constant is 0.0110s−1 . I got 32.4 degree C

1 answer:

gulaghasi [49]2 years ago

5 0

The rate constant is mathematically given as

K2=2.67sec^{-1}

<h3>What is the Arrhenius equation?</h3>

The rate constant for a particular reaction may be calculated with the use of the Arrhenius equation. This constant can be stated in terms of two distinct temperatures, T1 and T2, as follows:

$ln(\frac{K2}{K1})= (\frac{Ea}{R})*(\frac{1}{T1}-\frac{1}{T2})$

Therefore

KT1= 0.0110^{-1}

T1= 21+273.15

T1= 294.15K

T2= 200

T2=200+273.15

T2= 473.15K

Ea= 35.5 Kj/Mol

Hence, in j/mol R Ea is

Ea=35.5*1000 j/mol R

$ln(\frac{K2}{0.0110})= (\frac{35.5*1000}{8.314})*(\frac{1}{294.15}-\frac{1}{473.15}\\\\ln(\frac{K2}{0.0110})=5.492$

K2/0.0110 =e^(5.492)

K2/0.0110 =242.74

K2= 242.74*0.0110

K2=2.67sec^{-1}

In conclusion, rate constant

K2=2.67sec^{-1}

Read more about rate constant

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Be sure to answer all parts. For the titration of 10.0 mL of 0.250 M acetic acid with 0.200 M sodium hydroxide, determine the pH

DaniilM [7]

Explanation:

$Molarity=\frac{moles}{Volume(L)}$

Molarity of the acetic acid = 0.250 M

Volume of the acetic acid solution = 10.0 mL = 0.010 L( 1 mL =0.001L)

Moles of acetic acid ;

$n=0.250 M\times 0.010 L=0.0025 mol$

Molarity of the NaOH = 0.200 M

a) Volume of the NaOH solution = 10.0 mL = 0.010 L( 1 mL =0.001L)

Moles of NaOH : $0.200M\times 0.010 L=0.002 mol$

$CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O$

1 mole NaOH neutralizes 1 mole of acetic acid , then 0.002 moles of NaOH will neutralize 0.002 mol of acetic acid.

Moles of acetic acid left un-neutralized = 0.0025 mol - 0.002 = 0.0005 mol

1 mole of acetic acid gives 1 mole of hydrogen ion, then 0.0005 mole of acetic acid will give 0.0005 mole of hydrogen ions.

Moles of hydrogen ion= 0.0005 mol

Volume of the solution = 0.010 L+ 0.010 L = 0.020 L

$[H^+]=\frac{0.0005 mol}{0.020 L}=0.025 M$

The pH of the 10.0 mL of base added to acetic acid solution :

$pH=-\log[H^+]=-\log[0.025 M]=1.60$

b) Volume of the NaOH solution = 12.0 mL = 0.012 L( 1 mL =0.001L)

Moles of NaOH : $0.200M\times 0.012 L=0.0024 mol$

$CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O$

1 mole NaOH neutralizes 1 mole of acetic acid , then 0.0024 moles of NaOH will neutralize 0.0024 mol of acetic acid.

Moles of acetic acid left un-neutralized = 0.0025 mol - 0.0024 = 0.0001 mol

1 mole of acetic acid gives 1 mole of hydrogen ion, then 0.0001 mole of acetic acid will give 0.0001 mole of hydrogen ions.

Moles of hydrogen ion= 0.0001 mol

Volume of the solution = 0.010 L+ 0.012 L = 0.022 L

$[H^+]=\frac{0.0001 mol}{0.022 L}=0.0045 M$

The pH of the 12.0 mL of base added to acetic acid solution :

$pH=-\log[H^+]=-\log[0.0045 M]=2.34$

c) Volume of the NaOH solution = 15.0 mL = 0.015 L( 1 mL =0.001L)

Moles of NaOH : $0.200M\times 0.015 L=0.003 mol$

$CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O$

1 mole NaOH neutralizes 1 mole of acetic acid , then 0.003 moles of NaOH will neutralize 0.003 mol of acetic acid.

All the moles of acetic acid will get neutralized by NaOH and un-neutralized sodium hydroxide will left over.

Moles of NaOH left un-neutralized = 0.003 mol - 0.0025 = 0.0005 mol

1 mole of NaOH gives 1 mole of hydroxide ion, then 0.0005 mole of NaOH acid will give 0.0005 mole of hydroxide ions.

Moles of hydroxide ion= 0.0005 mol

Volume of the solution = 0.010 L+ 0.015 L = 0.025 L

$[OH^-]=\frac{0.0005 mol}{0.025 L}=0.02 M$

The pOH of the 15.0 mL of base added to acetic acid solution :

$pOH=-\log[OH^-]=-\log[0.02 M]=1.70$

The pH of the 15.0 mL of base added to acetic acid solution :

$pH=14-pOH=14-1.70=12.3$

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Murljashka [212]

Answer:

$31.8\ moles = 1.91 * 10^{25}\ molecules$

Explanation:

Given

31.8 moles of water

Required

Determine the number of molecule

In standard units: 1 molecule of water is:

$1\ mole = 6.022 * 10^{23}\ molecules$

Multiply both sides by 31.8

$31.8 * 1\ mole = 31.8 * 6.022 * 10^{23}\ molecules$

$31.8\ moles = 31.8 * 6.022 * 10^{23}\ molecules$

$31.8\ moles = 191.4996 * 10^{23}\ molecules$

Express 191.4996 as a standard form

$31.8\ moles = 1.914996 * 10^2 * 10^{23}\ molecules$

Apply law of indices:

$31.8\ moles = 1.914996 * 10^{2+23}\ molecules$

$31.8\ moles = 1.914996 * 10^{25}\ molecules$

Approximate:

$31.8\ moles = 1.91 * 10^{25}\ molecules$

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A tire is filled to a gauge pressure of 2 atm at 27 degrees Celsius after a drive the temperature within the tire rises to 47 de

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Answer:

2.13atm

Explanation:

The following information were obtained from the question:

P1 (initial pressure) = 2atm

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P2 (final pressure) =?

Using the equation P1/T1 = P2/T2, the final pressure within the tyre can be obtained as illustrated below:

P1/T1 = P2/T2

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