Ask question

Ask question

All categories

2 years ago

7

What is the activation energy of a reaction whose rate constant doubles when the temperature is increased from 300K to 350K (ass

uming that the activation energy and pre-exponential factor are both temperature independent)? Express your answer as a number with units.

1 answer:

bekas [8.4K]2 years ago

3 0

Answer:

12.1 kJ

Explanation:

For this problem, we are going to use the Arrhenius Equation.

$ln(\frac{k_{2} }{k_{1}})=\frac{E_{a} }{R} (\frac{1}{T_{1}} -\frac{1}{T_{2}} )\\ln(\frac{2}{1} )=\frac{E_{a} }{8.314} (\frac{1}{300}-\frac{1}{350} )$

We are going to solve for activation energy, $E_{a}$ .

$E_{a} =12.1$ kJ

You might be interested in

At Jim's auto shop, it takes him minutes to do an oil change and minutes to do a tire change. Let be the number of oil changes h

Juli2301 [7.4K]

Answer:

The total time that Jim needs to change x oil changes and y tire changes is less than 180 min.

The time needed for x oil changes is 12 * x.

The time needed for y tire changes is 18 * y.

The total time is the sum of the above times and needs to be less than 180 that is

12 * x + 18 * y < 180 divide both sides of equation by 6

12/6 * x + 18/6*y < 180/6

2*x + 3*y < 30

2*x < 30 - 3*y divide both sides by 2 to get the inequality for x

x < 30/2 - 3/2*y = 15 - 1.5 y < 15 that is x < = 15

2*x + 3*y < 30

3*y < 30 - 2*x divide both sides by 3 to get the inequality for y

y < 30/3 - 2/3 *x = 10 - 2/3*x < 10 that is y < = 10

Also we can write x + y < x+ 3/2 * y < 15.

Explanation:

Jim's can do not more then 5 oil changes and not more then 10 tire changes or all together she can do not more then 15 total of oil and tire changes.

5 0

3 years ago

Michelle is trying to find the average atomic mass of a sample of an unknown

GREYUIT [131]

The average atomic mass of her sample is 114.54 amu

Let the 1st isotope be A

Let the 2nd isotope be B

From the question given above, the following data were obtained:

Abundance of isotope A (A%) = 59.34%
Mass of isotope A = 113.6459 amu
Mass of isotope B = 115.8488 amu
Abundance of isotope B (B%) = 100 – 59.34 = 40.66%
Average atomic mass =?

The average atomic mass of the sample can be obtained as follow:

$Average \: atomic \: mass \: = \frac{mass \: of \: A \times A\%}{100} + \frac{mass \: of \: B \times B\%}{100} \\ \\ Average \: atomic \: mass \: = \frac{113.6459\times 59.34}{100} + \frac{115.8488\times 40.66}{100} \\ \\ Average \: atomic \: mass \: = 114.54 \: amu \\ \\$

Thus, the average atomic mass of the sample is 114.54 amu

Learn more about isotope: brainly.com/question/25868336

3 0

2 years ago

The acid HOCl (hypochlorous acid) is produced by bubbling chlorine gas through a suspension of solid mercury(II) oxide particles

Oliga [24]

Answer:

K = [ HOCl ] . [HgO. HgCl2] / [Cl2]^2 [H2O] [HgO]^2

Explanation:

The law of Mass Action states that, at constant temperature, the rate of reaction is proportional to the active masses of each of the reactants.

The reaction above is a reversible reaction and the law of mass action also applies to it.

The rate of reaction from left-to-right reaction = r1 = k. [Cl2]^2 [H2O] [HgO]^2

Rate of reaction from right - to - left r2 = k. [hocl]^2 [HgO . hgcl2]

Then at equilibrium,

r1 = r2

k1/k2 = [HOCl ]^2 [HgO. HgCl2] / [Cl2]^2 [H2O] [HgO]^2 = K

where K is the equilibrium constant for the reaction.

5 0

3 years ago

Cómo escribir el símbolo Theta

Pani-rosa [81]

Para escribir el símbolo theta, presione 03B8 + Alt + X.

8 0

2 years ago

Read 2 more answers

Practice entering numbers that include a power of 10 by entering the diameter of a hydrogen atom in its ground state, dH=1.06×10

kkurt [141]

Answer:

The diameter of the hydrogen $\mathbf{d =1.0605 \times 10^{-10}\ m}$

Explanation:

From the given information:

Using the concept of Bohr's Model, the equation for the angular momentum can be expressed as:

$L = \dfrac{nh}{2 \pi}$

Where the generic expression for angular momentum is:

L = mvr.

replacing the value of L into the previous equation, we have:

$mvr= \dfrac{nh}{2 \pi}$

$v= \dfrac{nh}{2 \pi mr}$ ----- (1)

The electron in the hydrogen atom posses an electrostatic force which gives a centripetal force.

$\dfrac{ke^2}{r^2} = \dfrac{mv^2}{r}$ ----- (2)

replacing the value of v in equation (1) into (2), and taking r as the subject of the formula, we have:

$\dfrac{ke^2}{r} = m (\dfrac{nh}{2 \pi mr})^2$

$ke^2=\dfrac{n^2h^2}{4 \pi^2 mr}$

$r =\dfrac{n^2h^2}{4 \pi^2 mke^2}$

For ground-state n = 1

$h = (6.625 \times 10^{-34} \ J.s)^2$

$m =( 9.1 \times 10^{-31} \ kg)(9 \times 10^9 \ N .m^2/C^2)$

$Ke = (1.6 \times 10^{-19} \ C)^2$

$r =\dfrac{(1)^2(6.625 \times 10^{-34})^2}{4 \pi^2 (9.1 \times 10^{-31} )(9 \times 10^9 ) (1.6 \times 10^{-19})^2}$

$r =\dfrac{4.3890625 \times 10^{-67}}{8.27720295 \times 10^{-57}}$

$\mathbf{r = 5.3025 \times 10^{-11} \ m}$

Therefore, the diameter of hydrogen d = 2r

$\mathbf{d = ( 2 \times 5.3025 \times 10^{-11} \ m})}$

$\mathbf{d =1.0605 \times 10^{-10}\ m}}$

4 0

3 years ago