13 × 10–5 mol L–1s–1
It may be remembered that
Rate
NO
=
[
]


1
4
2
∆
∆t
[
2]
NO
t
∆
=
∆
6 79 × 10–4 × 4 mol L–1 min–1 = 2 72 × 10–3 mol L–1min–1
Solution
Solution
Solution
Solution
Solution
3 2
3