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

Answers

Answer 1

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


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