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3 years ago

7

The second-order rate constants for the reaction of oxygen atoms ·with aromatic hydrocarbons have been measured (R. Atkinson and

J.N . Pitts, J. Phys. Chem. 79, 295 (1975)). In the reaction with benzene the rate constants are 1.44x 107dm' mor ' s ' at 300.3 K, 3.03 x 107 dm' mol"'s"' at 34 1.2K, and 6.9 x 107 dm' mor's ' at 392.2K. Find the frequency factor and activation energy of the reaction.

1 answer:

Blizzard [7]3 years ago

3 0

Answer: Frequency factor A = 8 × 10⁹

activation energy Ea = 15.5 KJ/Mol

Explanation: to begin, let us first define the parameters given;

K₁ = 1.44 × 10⁷dm³mol⁻¹s⁻¹

K₂ = 3.03 × 10⁷ dm³mol⁻¹s⁻¹

K₃ = 6.9 × 10 dm³mol⁻¹s⁻¹

also T₁ = 300.3 K

T₂ = 341.2 K

T₃ = 392.2 K

we know that;

㏑ K₂ / K₁ = Ea/R [1/T₁ -1/T₂]

where R is given as 8.314 J/mol-k

Ea = activation energy

K₁, K₂ = rate constant

T₁, T₂ = Temperature

therefore, ㏑ (3.03 × 10⁷/ 1.44 × 10⁷) = Ea / 8.314 [1/300.3 - 1/341.2]

this gives Ea = 15496.16 J/Mol ≈ 15.5 KJ /Mol

∴ Ea = 15.5 KJ/ Mol

also given that K = A e⁻∧Ea/RT

here A = frequency factor

∴ 6.9 × 10⁷ = A e⁻ ∧(15496.16/8.314 × 392.2)

A = 7.99 × 10⁹ = 8 × 10⁹

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Answer:

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Explanation:

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As $\Theta$ is very small so we can write $sin\Theta =tan\Theta$

So from eqn 1 and eqn 2 we can write

$y_1=\frac{1.22\lambda D}{d}$ --------eqn 3

Here $y_1$ is the position of first maxima D is the distance of screen from the circular aperture d is the diameter of aperture

It is given that diameter of circular aperture is 14.7 cm so $y_1=\frac{14.7}{2}=7.35 \ cm$

Now putting all these value in eqn 3

$d=\frac{1.22\lambda D}{y_1}$

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3 years ago

An infrared (IR) and ultraviolet (UV) wave propagating through a vacuum must have the same____

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Answer:

D. Wavelength

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An infrared (IR) and ultraviolet (UV) wave propagating through a vacuum must have the same wavelength.

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3 years ago

An object is going 22 m/s and is 3 kg. find momentum

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So what you wanna do is take your two givens. 22 and 3. Now you wanna take 22 and divide that by 3. And that gives you 7.33. Now if your answer HAS to be a whole number it'll be 66.

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A large airplane typically has three sets of wheels: one at the front and two farther back, one on each side under the wings. Co

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(a) The force the ground exerts on each set of rear wheels when the plane is at rest on the runway is 0.743 MN.

(b) The force the ground exerts on the front set of wheels is 0.239 MN.

<h3>Center mass of the airplane</h3>

The concept of center mass of an object can be used to dtermine the mass distribution of the airplane along the line through the center.

<h3>Some assumptions</h3>

The wheels under the wind do not pass through the center line.
The position of the front wheel is constant and it is zero mark (origin).
The rear wheels are at 21.7 m mark

Position of the center mass of the plane is calculated as follows;

Let the position of the center mass, Xcm = y

the center mass is 3 m in front of rear wheels, that is

21.7 - y = 3

y = 21.7 - 3

y = 18.7 m

Xcm = 18.7 m

<h3>Mass of the plane at the position of the rear wheels</h3>

Let the mass of the plane at front wheels = M1

Let the mass of the plane at rear wheels = M2

$X_{cm} = \frac{M_1x_1 + M_2x_2}{M_1 + M_2}$

$18.7 = \frac{M_1(0) + M_2(21.7)}{177000} \\\\3,309,900 = 21.7M_2\\\\M_2 = \frac{3,309,900}{21.7} \\\\M_2 = 152,529.95 \ kg$

<h3>Force exerted by the ground on each rear wheel</h3>

There are two rear wheels, and the force exerted on each wheel due to mass of the airplane at this position is calculated as follows;

$W = mg\\\\W_1 = W_2 = \frac{1}{2} (mg) = \frac{1}{2} (152,529.95 \times 9.8) = 743,396.76 \ N= 0.743 \ MN$

<h3>Mass of the plane at the position of the front wheel</h3>

M1 + M2 = 177,000

M1 = 177,000 - M2

M1 = 177,000 - 152,529.95

M1 = 24,470.05 kg

<h3>Force exerted by the ground on the front wheel</h3>

W = mg

W = 24,470.05 x 9.8

W = 239,806.5 N = 0.239 MN

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2 years ago

An elevator is moving down with an acceleration of 3.36 m/s2.

sergeinik [125]

Answer : 413.44N

Here it is given that an elevator is moving down with an acceleration of 3.36 m/s² . And we are interested in finding out the apparent weight of a 64.2 kg man . For the diagram refer to the attachment .

From the elevator's frame ( non inertial frame of reference) , we would have to think of a pseudo force.
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When a elevator accelerates down , the weight recorded is less than the actual weight .

From the Free body diagram ,

$\sf\longrightarrow Weight = mg - ma \\$

$\sf\longrightarrow Weight = m ( g - a ) \\$

Mass of the man = 64.2 kg

$\sf\longrightarrow Weight = 64.2( 9.8 - 3.36) N\\$

$\sf\longrightarrow Weight = 64.2 * 6.44 N\\$

$\sf\longrightarrow \underline{\boxed{\bf Weight_{apparent}= 413.44 N }} \\$

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2 years ago