For 4 NH₃ + 5 O₂ → 4 NO + 6 H₂O, if O₂ disappears at 0.50 M/s, NO appears at

0.40 M/s rate = (1/5)·d[O₂]/dt = (1/4)·d[NO]/dt → d[NO]/dt = (4/5)(0.50) = 0.40 M/s.

Reaction Rates

AP Chemistry· difficulty 3/5

For 4 NH₃ + 5 O₂ \to 4 NO + 6 H₂O, if O₂ disappears at 0.50 M/s, NO appears at

Explanation

rate = (1/5)\cdotd[O₂]/dt = (1/4)\cdotd[NO]/dt \to d[NO]/dt = (4/5)(0.50) = 0.40 M/s.

Want 10 more like this — adaptive to your weak spots?