The half-life for the first-order decomposition of is 1.3*10^-5.

N2O4--> 2NO2

If N2O4 is introduced into an evacuated flask at a pressure of 17.0 mmHg , how many seconds are required for the pressure of NO2 to reach 1.4mmHg ?

i know that the half life equation is t1/2= ln2/k which allowed me to calculate k, but i am having trouble with figuring out how the stoichiometry and pressure come into play. Since there are 2 moles of NO2 and only one mole of N2O4.

2 answers

The question doesn't ask anything about NO2; only N2O4. I would ignore the NO2 since the question is about pressure of N2O4 and it will exhibit its own partial pressure independent of NO2.
did you ever solve this?? because i need help on it too
Similar Questions
    1. answers icon 2 answers
  1. Dinitrogen tetroxide decomposes to nitrogen dioxide:N2O4 (g) ---> 2NO2 (g) Delta H rxn: 55.3 kJ At 298 K a reaction vessel
    1. answers icon 1 answer
  2. 92.01 grams of N2O4 (g) is placed in a container and allowed to dissociate.N2O4 (g) --> 2NO2 (g) The mixture of N2O4 and NO2
    1. answers icon 2 answers
  3. Dinitrogen tetroxide decomposes to nitrogen dioxide:N2O4(g)→2NO2(g)ΔHorxn=55.3kJ At 298 K, a reaction vessel initially
    1. answers icon 1 answer
more similar questions