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aniked [119]
3 years ago
11

The rate constant for a first order reaction

Chemistry
1 answer:
Lena [83]3 years ago
8 0

Answer:

E_a = 103.626 × 10³ KJ/mol

Explanation:

Formula to solve this is given by;

Log(k2/k1) = (E_a/2.303R)((1/T1) - (1/T2))

Where;

k2 is rate constant at second temperature

k1 is rate constant at first temperature

R is universal gas constant

T1 is first temperature

T2 is second temperature

We are given;

k1 = 2.8 × 10^(-3) /s

k2 = 4.8 × 10^(-4) /s

R = 8.314 J/mol.k

T1 = 60°C = 333.15 K

T2 = 45°C = 318.15 K

Thus;

Log((4.8 × 10^(-4))/(2.8 × 10^(-3))) = (E_a/(2.303 × 8.314))((1/333.15) - (1/318.15))

We now have;

-0.76592 = -0.00000739121E_a

E_a = -0.76592/-0.00000739121

E_a = 103.626 × 10³ KJ/mol

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2 years ago
discuss how variations in electronegativity result in the unequal sharing of electrons in polar molecules.
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Variations in electronegativity prompt in the unequal halves of electrons in polar molecules because when one atom is more electronegative than the other, it becomes more polar than the other.

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Polar molecules have unequal sharing of electrons because the atoms have unequal attraction for electrons so the sharing is unequal.

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Answer:

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3 years ago
A mixture of noble gases [helium (MW 4), argon (MW 40), krypton (MW 83.8), and xenon (MW 131.3)] is at a total pressure of 150 k
kodGreya [7K]

Answer:

a) 1,6%

b) 64,775 g/mol

c) 3,6×10⁻² M

d) 2,3×10⁻³ g/mL

Explanation:

a) The mass fractium of helium is obtained converting the moles of the four gases to grams with molar weight and then caculating of the total of grams how many are of helium, thus:

  • Helium: 0,25 moles ×\frac{4 g}{1 mol} = 1 g of Helium
  • Argon: 0,25 moles ×\frac{40 g}{1 mol} = 10 g of Argon
  • Krypton: 0,25 moles ×\frac{83,8 g}{1 mol} = 20,95 g of krypton
  • Xenon: 0,25 moles ×\frac{131,3 g}{1 mol} = 32,825 g of Xenon

Total grams: 1g+10g+20,85g+30,825g= 62,675 g

Mass fraction of helium: \frac{1 gHelium}{62,675 g} × 100 = <em>1,6%</em>

<em />

<em>The mass fraction of Helium is 1,6%</em>

<em />

<em>b)</em><em>  </em>Because the mole fraction of all gases is the same the average molecular weight of the mixture is:

\frac{4+40+83,8+131,3}{4} = 64,775 g/mol

c) The molar concentration is possible to know ussing ideal gas law, thus:

\frac{P}{R.T} = M

Where:

P is pressure: 150 kPa

R is gas constant: 8,3145\frac{L.kPa}{K.mol}

T is temperature: 500 K

And M is molar concentration. Replacing:

M = 3,6×10⁻² M

d) The mass density is possible to know converting the moles of molarity to grams with average molecular weight and liters to mililiters, thus:

3,6×10⁻² \frac{mol}{L} × \frac{64,775 g}{mol} × \frac{1L}{1000 mL} =

2,3×10⁻³ g/mL

I hope it helps!

7 0
3 years ago
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