N2+3H2 →2NH3. During the formation of ammonia 2 moles of nitrogen disappear in 1min. Find (i) ROR (ii) ROD H2 (iii) ROA of NH3. The volume of vessel is 1L.
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7
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.
Answered by
2
Answer:
ROR = 0.03 mol/L sec.
ROD)H2 = 0.09 mol/L sec.
ROD)NH3 = 0.06 mol/L sec.
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