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taurus [48]
2 years ago
12

Calculate the kinetic energy of co2 at 254 k .

Chemistry
1 answer:
san4es73 [151]2 years ago
3 0
You need to take into account the equations for the average speed and kinetic energy of gases.

Kinetic energy, KE = [1/2] m * v^2

Average speed, v = √[3RT/m] => v^2 = 3RT / m

=> KE = [1/2]m * 3RT/m = [3/2] RT

R = 8,314 J/mol*K

T = 254 K

=> KE = [3/2]*(8,314J/mol*K)*(254K) = 3167.6 J/mol

Answer: 3167.6 J/mol
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What is the volume of 14.0g of nitrogen gas at STP?
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Answer:

  • <em>The volume of 14.0 g of nitrogen gas at STP is </em><u><em>11.2 liter.</em></u>

Explanation:

STP stands for standard pressure and temperature.

The International Institute of of Pure and Applied Chemistry, IUPAC changed the definition of standard temperature and pressure (STP) in 1982:

  •   Before the change, STP was defined as a temperature of 273.15 K and an absolute pressure of exactly 1 atm (101.325 kPa).

  •    After the change, STP is defined as a temperature of 273.15 K and an absolute pressure of exactly 105 Pa (100 kPa, 1 bar).

Using the ideal gas equation of state, PV = nRT you can calculate the volume of one mole (n = 1)  of gas. With the former definition, the volume of a mol of gas at STP, rounded to 3 significant figures, was 22.4 liter. This is classical well known result.

With the later definition, the volume of a mol of gas at STP is 22.7 liter.

I will use the traditional measure of 22.4 liter per mole of gas.

<u>1) Convert 14.0 g of nitrogen gas to number of moles:</u>

  • n = mass in grams / molar mass
  • Atomic mass of nitrogen: 14.0 g/mol
  • Nitrogen gas is a diatomic molecule, so the molar mass of nitrogen gas = molar mass of N₂ = 14.0 × 2 g/mol = 28.0 g/mol
  • n = 14.0 g / 28.0 g/mol = 0.500 mol

<u>2) Set a proportion to calculate the volume of nitrogen gas:</u>

  • 22.4 liter / mol = x  / 0.500 mol
  • Solve for x: x = 0.500 mol × 22.4 liter / mol = 11.2 liter.

<u>Conclusion:</u> the volume of 14.0 g of nitrogen gas at STP is 11.2 liter.

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Answer: II. The catalyst provides a new pathway in the reaction mechanism.

III. The catalyst speeds up the reaction.

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