Chemistry Exam Notes: Pressure and Boiling Point

Q34 Pressure

  • Reaction: N<em>2+3F</em>22NF3N<em>2 + 3F</em>2 \rightarrow 2NF_3
  • Initial Partial Pressures:
    • N2N_2: 1 atm
    • F2F_2: 1.8 atm
  • Reaction Completion: The reaction goes to completion, and VV (volume) and TT (temperature) remain constant.
  • Objective: Determine the maximum partial pressure of NF3NF_3 produced.
  • Relationship: Since volume (V) and temperature (T) are constant, the number of moles is directly proportional to the partial pressure.

Calculations

  • Based on N2N_2:
    • 1 atm N<em>2N<em>2 can produce: 1 atm N</em>2×2 NF<em>31 N</em>2=2.0 atm NF31 \text{ atm } N</em>2 \times \frac{2 \ NF<em>3}{1 \ N</em>2} = 2.0 \text{ atm } NF_3
  • Based on F2F_2:
    • 1.8 atm F<em>2F<em>2 can produce: 1.8 atm F</em>2×2 NF<em>33 F</em>2=1.2 atm NF31.8 \text{ atm } F</em>2 \times \frac{2 \ NF<em>3}{3 \ F</em>2} = 1.2 \text{ atm } NF_3
  • Limiting Reactant: F<em>2F<em>2 is the limiting reactant because it produces less NF</em>3NF</em>3.
  • Maximum Partial Pressure of NF3NF_3: 1.20 atm

Q37 Boiling Point

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