Answer to Question #108427 in General Chemistry for Zac

Question

For the equilibrium reaction: 2NH3(g) ⇌ N2(g) + 3H2(g) ∆H = +37.2kJ
[N2] = 0.25M; [H2] = 0.28M, and [NH3] = 18.1M.
The rate of loss of Ammonia gas is 0.1 molL-1s-1, what is the rate of production of Nitrogen?

Answer 1

Rate of reaction is d[N₂]/dt = - d[NH₃]/2dt, so if -d[NH₃]/dt = 0.1 mol/(L*s) the rate of nitrogen production is 0.1/2 = 0.05 mol/(L*s)