At what temperature will water change from a liquid to a solid?

A certain first-order reaction has a half-life of 25.2 s at 20°C. What is the value of the rate constant k at 60°C if the activa

DochEvi [55]

Answer:

t

(

2

)

1/2

=

85.25 s

Notice how you're given the half-life (for one temperature), a second temperature, and the activation energy. The key to doing this problem is recognizing that:

the half-life for a first-order reaction is related to its rate constant.

the rate constant changes at different temperatures.

Go here for a derivation of the half-life of a first-order reaction. You should find that:

t

1/2

=

ln

2

k

Therefore, if we label each rate constant, we have:

k

1

=

ln

2

t

(

1

)

1/2

k

2

=

ln

2

t

(

2

)

1/2

Recall that the activation energy can be found in the Arrhenius equation:

k

=

A

e

−

E

a

/

R

T

where:

A

is the frequency factor, i.e. it is proportional to the number of collisions occurring over time.

E

a

is the activation energy in

kJ/mol

.

R

=

0.008314472 kJ/mol

⋅

K

is the universal gas constant. Make sure you get the units correct on this!

T

is the temperature in

K

(not

∘

C

).

Now, we can derive the Arrhenius equation in its two-point form. Given:

k

2

=

A

e

−

E

a

/

R

T

2

k

1

=

A

e

−

E

a

/

R

T

1

we can divide these:

k

2

k

1

=

e

−

E

a

/

R

T

2

e

−

E

a

/

R

T

1

Take the

ln

of both sides:

ln

(

k

2

k

1

)

=

ln

(

e

−

E

a

/

R

T

2

e

−

E

a

/

R

T

1

)

=

ln

(

e

−

E

a

/

R

T

2

)

−

ln

(

e

−

E

a

/

R

T

1

)

=

−

E

a

R

T

2

−

(

−

E

a

R

T

1

)

=

−

E

a

R

[

1

T

2

−

1

T

1

]

Now if we plug in the rate constants in terms of the half-lives, we have:

ln

⎛

⎜

⎝

ln

2

/

t

(

2

)

1/2

ln

2

/

t

(

1

)

1/2

⎞

⎟

⎠

=

−

E

a

R

[

1

T

2

−

1

T

1

]

This gives us a new expression relating the half-lives to the temperature:

⇒

ln

⎛

⎜

⎝

t

(

1

)

1/2

t

(

2

)

1/2

⎞

⎟

⎠

=

−

E

a

R

[

1

T

2

−

1

T

1

]

Now, we can solve for the new half-life,

t

(

2

)

1/2

, at the new temperature,

40

∘

C

. First, convert the temperatures to

K

:

T

1

=

25

+

273.15

=

298.15 K

T

2

=

40

+

273.15

=

313.15 K

Finally, plug in and solve. We should recall that

ln

(

a

b

)

=

−

ln

(

b

a

)

, so the negative cancels out if we flip the

ln

argument.

⇒

ln

⎛

⎜

⎝

t

(

2

)

1/2

t

(

1

)

1/2

⎞

⎟

⎠

=

E

a

R

[

1

T

2

−

1

T

1

]

⇒

ln

⎛

⎜

⎝

t

(

2

)

1/2

400 s

⎞

⎟

⎠

=

80 kJ/mol

0.008314472 kJ/mol

⋅

K

[

1

313.15 K

−

1

298.15 K

]

=

(

9621.78 K

)

(

−

1.607

×

10

−

4

K

−

1

)

=

−

1.546

Now, exponentiate both sides to get:

t

(

2

)

1/2

400 s

=

e

−

1.546

⇒

t

(

2

)

1/2

=

(

400 s

)

(

e

−

1.546

)

=

85.25 s

This should make sense, physically. From the Arrhenius equation, the higher

T

2

is, the more negative the

[

1

T

2

−

1

T

1

]

term, which means the larger the right hand side of the equation is.

The larger the right hand side gets, the larger

k

2

is, relative to

k

1

(i.e. if

ln

(

k

2

k

1

)

is very large,

k

2

>>

k

1

). Therefore, higher temperatures means larger rate constants.

Furthermore, the rate constant is proportional to the rate of reaction

r

(

t

)

in the rate law. Therefore...

The higher the rate constant, the faster the reaction, and thus the shorter its half-life should be.

Explanation:

Sorry just go here https://socratic.org/questions/588d14f211ef6b4912374c92#370588

3 0

2 years ago

CH3CH2OH can interact with<br> other like molecules through ___?

kkurt [141]

Answer: dispersion forces, dipole-dipole and hydrogen bonds

Explanation:

4 0

2 years ago

Match the prefix used in naming binary molecule compounds with its corresponding number

VARVARA [1.3K]

Nona= 9, hepta= 7, hexa= 6, tetra= 4

8 0

2 years ago

The following equation for the reaction of alum with barium chloride is not balanced. KAl(SO4)2•12H2O(aq) + BaCl2(s) → KCl(aq) +

Pepsi [2]

Answer:

A. 1.04g of BaCl2.

B. Percentage yield of BaSO4 is 87.6%

Explanation:

A. The balanced equation for the reaction. This given below:

KAl(SO4)2•12H2O(aq) + 2BaCl2(s) → KCl(aq) + AlCl3(aq) + 2BaSO4(s) + 12H2O(l)

Next, we shall determine the number of mole in 25 mL of a 0.10 M alum. This is illustrated below:

Volume = 25mL = 25/1000 = 0.025L

Molarity = 0.1M

Mole =..?

Mole = Molarity x Volume

Mole of alum = 0.1 x 0.025 = 2.5x10¯³ mol.

Next, we shall convert 2.5x10¯³ mol of alum to grams.

Number of mole alum, KAl(SO4)2•12H2O = 2.5x10¯³ mol

Molar Mass of alum, KAl(SO4)2•12H2O = 39 + 27 + 2[32+(16x4)] + 12[(2x1) + 16]

= 39 + 27 + 2[32 + 64] + 12[2 + 16]

= 39 + 27 + 2[96] + 12[18]

= 474g/mol

Mass of alum, KAl(SO4)2•12H2O =..?

Mass = mole x molar mass

Mass of alum, KAl(SO4)2•12H2O = 2.5x10¯³ x 474 = 1.185g

Next, we shall determine the mass of alum and BaCl2 that reacted and the mass of BaSO4 produced from the balanced equation. This is illustrated below:

Molar mass of alum, KAl(SO4)2•12H2O = 474g

Mass of alum, KAl(SO4)2•12H2O from the balanced equation = 1 x 474 = 474g

Molar mass of BaCl2 = 137 + (35.5x2) = 208g/mol

Mass of BaCl2 from the balanced equation = 2 x 208 = 416g

Molar mass of BaSO4 = 137 + 32 + (16x4) = 233g/mol

Mass of BaSO4 from the balanced equation = 2 x 233 = 466g

Summary:

From the balanced equation above,

474g of alum reacted with 416g of BaCl2 to produce 466g of BaSO4.

Finally, we can calculate the mass of BaCl2 needed for the reaction as follow:

From the balanced equation above,

474g of alum reacted with 416g of BaCl2.

Therefore, 1.185g of alum will react with = (1.185 x 416)/474 = 1.04g of BaCl2.

Therefore, 1.04g of BaCl2 is needed for the reaction.

B. Determination of the percentage yield of BaSO4(s).

We'll begin by calculating the theoretical yield of BaSO4. This is illustrated below:

From the balanced equation above,

474g of alum reacted to produce 466g of BaSO4.

Therefore, 1.185g of alum will react to produce = (1.185 x 466)/474 = 1.165g of BaSO4.

Therefore, the theoretical yield of BaSO4 is 1.165g.

Finally, we shall determine the percentage of BaSO4 as follow:

Actual yield of BaSO4 = 1.02g.

Theoretical yield of BaSO4 = 1.165g.

Percentage yield of BaSO4 =..?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield of BaSO4 = 1.02/1.165 x 100

Percentage yield of BaSO4 = 87.6%

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2 years ago

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when wax is heated, it turns into a liquid. how are the wax molecules affected by this change of state?

Vladimir [108]

Wax molecules have most of their weak van der waals interactions and other hydrophobic interactions decrease and thus most of them are broken between these lipid substances, allowing for the increased molecular movement of these substances. As there are fewer interactions between similar compounds.

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