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Problem

Answer published by the source. Consult the official source to check your work against its answer.

question

The vapor pressure of ethylene is given as a function of temperature by the equation log P = - [(834.13) / T] + 1.75 log T - 8.375 × 10^-3T + 5.32340 Calculate the enthalpy of vaporization of ethylene at its normal boiling point (- 103.9°C).

category

chemistry

options

  1. 12.838 kJ mol^-1
  2. 16.838 kJ mol^-1
  3. 18.838 kJ mol^-1
  4. 19.838 kJ mol^-1
  5. 13.838 kJ mol^-1
  6. 11.838 kJ mol^-1
  7. 15.838 kJ mol^-1
  8. 14.838 kJ mol^-1
  9. 10.838 kJ mol^-1
  10. 17.838 kJ mol^-1

src

stemez-PhysicalChemistry

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Official source

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