Convert phenol to anisole

Calculate the equilibrium constant k for the isomerization of glucose-1-phosphate to fructose-6-phosphate at 298 k. express your

k0ka [10]

We cannot solve this problem without using empirical data. These reactions have already been experimented by scientists. The standard Gibb's free energy, ΔG°, (occurring in standard temperature of 298 Kelvin) are already reported in various literature. These are the known ΔG° for the appropriate reactions.

<span>glucose-1-phosphate⟶glucose-6-phosphate ΔG∘=−7.28 kJ/mol
fructose-6-phosphate⟶glucose-6-phosphate ΔG∘=−1.67 kJ/mol
</span>
Therefore, the reaction is a two-step process wherein glucose-6-phosphate is the intermediate product.

glucose-1-phosphate⟶glucose-6-phosphate⟶fructose-6-phosphate

In this case, you simply add the ΔG°. However, since we need the reverse of the second reaction to end up with the terminal product, fructose-6-phosphate, you'll have to take the opposite sign of ΔG°.

ΔG°,total = −7.28 kJ/mol + 1.67 kJ/mol = -5.61 kJ/mol

Then, the equation to relate ΔG° to the equilibrium constant K is

ΔG° = -RTlnK, where R is the gas constant equal to 0.008317 kJ/mol-K.
-5.61 kJ./mol = -(0.008317 kJ/mol-K)(298 K)(lnK)
lnK = 2.2635
K = e^2.2635
K = 9.62

6 0

3 years ago

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A first-order reaction is 45% complete at the end of 43 minutes. What is the length of the half-life of this reaction

shtirl [24]

The half-life of the reaction is 50 minutes

Data;

Time = 43 minutes
Type of reaction = first order
Amount of Completion = 45%

<h3>Reaction Constant</h3>

Let the initial concentration of the reaction be X $_0$

The reactant left = (1 - 0.45) X $_0$ = 0.55 X $_0$ = X

For a first order reaction

$\ln(\frac{x}{x_o}) = -kt\\ k = \frac{1}{t}\ln (\frac{x_o}{x}) \\ k = \frac{1}{43}\ln (\frac{x_o}{0.55_o})\\ k = 0.013903 min^-^1$

The half-life of a reaction is said to be the time required for the initial amount of the reactant to reach half it's original size.

$x = \frac{x_o}{2} \\t = t_\frac{1}{2} \\t_\frac{1}{2} = \frac{1}{k}\ln(\frac{x_o}{x_o/2})\\$

Substitute the values

$t_\frac{1}{2} = \frac{1}{k}\ln(2)=\frac{0.6931}{0.013903}\\t_\frac{1}{2}= 49.85 min = 50 min$

The half-life of the reaction is 50 minutes

Learn more on half-life of a first order reaction here;

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4 0

2 years ago

What is the standard enthalpy of a reaction for which the equilibrium constant is (a) doubled, (b) halved when the temperature i

Alexxandr [17]

Answer:

a) 48KJ

b) -48KJ

Explanation:

Given that;

ln(K2/K1) = ΔH°/R(1/T2 - 1/T1)

K2= equilibrium constant at T2

K1 = equilibrum constant at T1

R = gas constant

T1 = initial temperature

T2 = final temperature

When we double the equilibrium constant K1; K2 = 2K1

T1 = 310 K

T2 = 310 + 15 = 325 K

ln(2K1/K1) =- ΔH°/R(1/T2 - 1/T1)

ln2 = -ΔH°/8.314(1/325 - 1/310)

0.693 = -ΔH°/8.314(3.08 * 10^-3 - 3.2 * 10^-3)

0.693 = -ΔH°/8.314 (-0.00012)

0.693 = 0.00012ΔH°/8.314

0.693 * 8.314 = 0.00012ΔH°

ΔH° = 0.693 * 8.314/0.00012

ΔH° = 48KJ

b) K2 =K1/2

ln(K1/2/K1) =- ΔH°/R(1/T2 - 1/T1)

ln (1/2) = -ΔH°/8.314 (1/325 - 1/310)

-0.693 = -ΔH°/8.314 (-0.00012)

-0.693 = 0.00012ΔH°/8.314

-0.693 * 8.314 = 0.00012ΔH°

ΔH°= -0.693 * 8.314/0.00012

ΔH°= -48KJ

6 0

3 years ago

A student has a sample of 1.33 moles of fluorine gas that is contained in a 23.3 L container at 314 K. What is the pressure of t

Gelneren [198K]

1.4715 atm is the pressure of the sample 1.33 moles of fluorine gas that is contained in a 23.3 L container at 314 K.

What is an ideal equation?

The ideal gas equation, pV = nRT, is an equation used to calculate either the pressure, volume, temperature or number of moles of a gas. The terms are: p = pressure, in pascals (Pa).

Given data:

Volume (V) = 23.3 L

Number of mole (n) = 1.33 moles

Temperature (T) = 314 K

Gas constant (R) = 0.821 atm.L/Kmol

Pressure (P) =?

The pressure inside the container can be obtained by using the ideal gas equation as illustrated below:

PV = nRT

P × 23.3 L = 1.33 moles × 0.0821 ×314 K

P = 1.4715 atm

Therefore, the pressure of the sample is 1.4715 atm.

Learn more about the ideal gas equation:

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SPJ1

4 0

2 years ago

Which lists only abiotic factors in this environment?

alukav5142 [94]

In this case, among the list of components given, the abiotic factors are water, rock, soil, and sun.

<h3>What is an abiotic factor?</h3>

An abiotic factor is a non-living part of an ecosystem that shapes its environment.

Biotic and abiotic factors make up a community via interaction.

Biotic factors are considered living things (having "life") while abiotic factors are simply non-living things.

Hence, in this case, among the list of components given, the abiotic factors are water, rock, soil, sun

Learn more about the abiotic factor here:

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4 0

2 years ago