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solong [7]
3 years ago
10

The first-order rearrangement of CH3NC is measured to have a rate constant of 3.61 × 10–15 s–1 at 298 K and a rate constant of 8

.66 × 10–7 s–1 at 425 K. Determine the activation energy for this reaction.
Physics
1 answer:
netineya [11]3 years ago
7 0

Answer:

The activation energy for this reaction, Ea = 159.98 kJ/mol

Explanation:

Using the Arrhenius equation as:

ln\frac {K_2}{K_1}=-\frac {E_a}{R}\times (\frac {1}{T_2}-\frac {1}{T_1})

Where, Ea is the activation energy.

R is the gas constant having value 8.314 J/K.mol

K₂ and K₁ are the rate constants

T₂ and T₁ are the temperature values in kelvin.

Given:

K₂ = 8.66×10⁻⁷ s⁻¹ , T₂ = 425 K

K₁ = 3.61×10⁻¹⁵ s⁻¹ , T₁ = 298 K

Applying in the equation as:

ln\frac {8.66\times 10^{-7}}{3.61\times 10^{-15}}=-\frac {E_a}{8.314}\times (\frac {1}{425}-\frac {1}{298})

Solving for Ea as:

Ea = 159982.23 J /mol

1 J/mol = 10⁻³ kJ/mol

Ea = 159.98 kJ/mol

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Answer:

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Explanation:

From the question given, Gas 1 was initially at a higher temperature than Gas 2.

As the two gas mixes together, there will be a transfer of heat from Gas 1 molecules to Gas 2 molecules. Now, as this continues over a period of time, the two gas will eventually have the same temperature.

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3 years ago
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The cornea behaves as a thin lens of focal lengthapproximately 1.80 {\rm cm}, although this varies a bit. The material of whichi
Keith_Richards [23]

Answer:

Explanation:

  a )

from lens makers formula

\frac{1}{f} =(\mu-1)(\frac{1}{r_1} -\frac{1}{r_2})

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putting the values

\frac{1}{1.8} =(1.38-1)(\frac{1}{.5} -\frac{1}{r_2})

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1 / r₂ = .538

r₂ = 1.86 cm .

= 18.6 mm .

b )

object distance u = 25 cm

focal length of convex lens  f  = 1.8 cm

image distance  v   = ?

lens formula

\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

\frac{1}{v} - \frac{1}{-25} = \frac{1}{1.8}

\frac{1}{v} = \frac{1}{1.8} -\frac{1}{25}

.5555 - .04

= .515

v = 1.94 cm

c )

magnification = v / u

= 1.94 / 25

= .0776

size of image = .0776 x size of object

= .0776 x 10 mm

= .776 mm

It will be a real image and it will be inverted.

 

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3 years ago
Compare the strengths of UV light and microwaves (in Hz). Which type of light is more powerful and how do you know?
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Answer:

UV light is more powerful as it has greater energy.

Explanation:

The energy propagated by electromagnetic waves ( light ) through vacuum or medium is known as electromagnetic radiation.

The frequency/wavelength range of electromagnetic radiation is known as electromagnetic spectrum. The electromagnetic spectrum ranging from gamma ray to radio waves.

Frequency range of UV light = ( 8 x 10¹⁴ to 3 x 10¹⁶ ) Hz

Frequency range of Microwaves = ( 300 x 10⁶ to 300 x 10⁹ ) Hz

Ratio of UV light to Microwaves = (\frac{8\times10^{14} }{300\times10^{6} } to \frac{3\times10^{16} }{300\times10^{9} } )

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Energy of electromagnetic radiation is given by the relation:

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UV light is more powerful than Microwaves as frequency of UV light is greater than frequency of microwaves. Thus, by the above equation, the energy of UV light is more than energy of Microwaves.  

5 0
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The smallest known galaxy, Segue 2, has an approximate radius of 1.05 × 1015 kilometers. Use the conversion factors 1 light-year
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Answer:

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