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N2 + 3h2 - 2Nh3...During the …

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N2 + 3h2 - 2Nh3...During the formation of Ammonia 2 mol of N2 disappear

Posted by Mayank Tiwari 4 years, 1 month ago

CBSE > Class 12 > Chemistry

1 answers

Gaurav Seth 4 years, 1 month ago

**Rate of change of concentration ammonia = -0.2 * 10^-4 M/s**

Explanation:

Given: N2+3H2 →2NH3. During the formation of ammonia 2 moles of nitrogen disappear in 1min. The volume of vessel is 1L.

Solution:

N2 + 3H2 -----> 2NH3

-d[H2] / dt = 0.3 * 10^-4 M/s

Rate = (1/3)(-d[H2] / dt)

= (1/2) (d[NH3] /dt)

= d[NH3] / dt

= (2/3)( -d[h2]/dt)

= d [NH3]/dt

d[NH3] /dt = -2/3 (0.3 * 10^-4)

d[NH3] /dt = -0.2 * 10^-4 M/s

Rate of change of concentration ammonia is the answer.

0Thank You

ANSWER

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