N2 + 3h2 - 2Nh3...During the …
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Gaurav Seth 5 years, 2 months 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.
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