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.
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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.
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