What is the average translational kinetic energy of molecules in a gas at temperature 27°C?

Correct answer: C. 6.21 x 10^-21 J

  • A. 3.23 x 10^-21 J
  • B. 4.11 x 10^-21 J
  • C. 6.21 x 10^-21 J
  • D. 7.71 x 10^-21 J
  • E. 9.11 x 10^-21 J

Explanation

The average translational kinetic energy of molecules in a gas is given by the formula: K.E. = (3/2) kTWhere K.E. is the average translational kinetic energy, k is the Boltzmann constant (approximately 1.38 x 10^-23 J/K), and T is the temperature in Kelvin.The given temperature is 27°C, which must be converted to Kelvin by adding 273.15:T = 27°C + 273.15 = 300.15 KSubstituting these values into the formula gives:K.E. = (3/2) * (1.38 x 10^-23 J/K) * (300.15 K)= 6.21 x 10-21 JTherefore, the correct answer is 6.21 x 10-21 J. The other options result from errors in calculation or incorrect conversions.

Written and checked by , M.Phil ChemistryLast updated
Report an error

The more specific you are, the faster it gets fixed. A source beats an opinion.

Prefer email? [email protected]

About Reaction Kinetics

Reaction kinetics relates reaction rate to concentration, temperature, surface area and catalysts. Questions cover rate laws, reaction order, rate constants, activation energy and the activated complex, including how a catalyst lowers the activation energy without changing the overall energy change or equilibrium position.

Practise Reaction Kinetics

573 free Reaction Kinetics MCQs from Chemistry, each with the correct answer and an explanation. Unlimited attempts, no account needed.

Discussion

Stuck on an option, or know a faster way to get there? Ask or explain it here.

No comments yet. Be the first to explain this one.

Exams that ask Chemistry questions like this

Chemistry is on 14 papers prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for every one of them.

More Reaction Kinetics questions