Question

If Δ[NO]/ Δt = –20.0 M/s under a given set of conditions, what are the rates of change of [N2] and [H2O]?

Answer 1

Answer:

the rate of change of $[N_2]$ = 10.0 M/s

the rate of change of  $[H_2O]$ = 20.0 M/s

Explanation:

Given that $\frac{d[NO]}{dt}$ = 20.0 M/s

The equation for the reaction can be written as:

$2 NO_{(g)} + 2H_2--->2H_2}O{(g)}+N_2_{(g)}$

The rate of change of $[N_2]$ can be calculated as:

$\frac{-1}{2} \frac{d[NO]}{dt} = \frac{d[N_2]}{dt}$

$\frac{-1}{2} (-20.0M/s)= \frac{d[N_2]}{dt}$

$-1 *-10.0M/s= \frac{d[N_2]}{dt}$

$10.0M/s= \frac{d[N_2]}{dt}$

$\frac{d[N_2]}{dt}= 10.0M/s$

∴ the rate of change of $[N_2]$ = 10.0 M/s

The rate of change of $[H_2O]$ can be calculated as:

$\frac{-1}{2} \frac{d[NO]}{dt} = \frac{+1}{2} \frac{d[H_2O]}{dt}$

$\frac{-1}{2} (-20.0M/s)= \frac{+1}{2} \frac{d[H_2O]}{dt}$

$10.0 M/s= \frac{+1}{2} \frac{d[H_2O]}{dt}$

$10.0 M/s*2= \frac{d[H_2O]}{dt}$

$\frac{d[H_2O]}{dt} = 20.0 M/s$

∴ the rate of change of  $[H_2O]$ = 20.0 M/s