Question

N2 + 3H2 2NH3

If N2 is reacting at a rate of –0.40 M/s, what is the rate by which NH3 is being produced? (Also, show calculations.)

-0.80 M/s
+0.40 M/s
+0.80 M/s
+1.20 M/s

Answer 1

+0.8M/s

Step-by-step explanation:

We are given that

`N_2+3H_2\rightarrow 2NH_3`

Rate of reacting of N2, d/dt([N2])=-0.4M/s

We have to find the rate by which NH3 is being produced.

We know that

`-(d[N_2])/(dt)=(1)/(2)(d[NH_3])/(dt)`

Using the formula

`-(-0.4)=(1)/(2)(d[NH_3])/(dt)`

`(d[NH_3])/(dt)=0.4* 2`

`(d[NH_3])/(dt)=0.8M/s`

Hence, the rate by which NH3 is being produced=0.8M/s

Option C is correct.

0.8M/s