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For the synthesis of ammonia from H2(g) and N2(g) by the reaction below, K = 6.2x10^5 at 298 K and K = 9.06x10^-2 at 500
For the synthesis of ammonia from H2(g) and N2(g) by the reaction below, K = 6.2x10^5 at 298 K and K = 9.06x10^-2 at 500 K.
3H2(g) + N2(g) 2NH3(g)
The reaction is exothermic by 92.23 kJ/mole. Suppose that hydrogen and nitrogen are reacted in the proper 3:1 stoichiometric ratio at 500K. If the initial pressure is 1 atm, what is the % yield of ammonia?
This is the work I have done on Maple -- but the concentration of ammonia seems very low to me. Also-- how would I calculate the %yield once I obtain the concentration of Ammonia?
Thanks!
5.24109=Kc(RT)(n) :=2 Kp:=5.242858645109 P(H2):=0.75atm P(N2):=0.25atm PV=nRT P=cRTStep by Step Solution
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