Answer:
+0.8M/s
Explanation:
We are given that
[tex]N_2+3H_2\rightarrow 2NH_3[/tex]
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
[tex]-\frac{d[N_2]}{dt}=\frac{1}{2}\frac{d[NH_3]}{dt}[/tex]
Using the formula
[tex]-(-0.4)=\frac{1}{2}\frac{d[NH_3]}{dt}[/tex]
[tex]\frac{d[NH_3]}{dt}=0.4\times 2[/tex]
[tex]\frac{d[NH_3]}{dt}=0.8M/s[/tex]
Hence, the rate by which NH3 is being produced=0.8M/s
Option C is correct.
0.8M/s
