Problem: Ammonia can be synthesized according to the following reaction: N 2(g) + 3H2(g) ⇌ 2NH3(g); Kp = 5.3 x 10–5 at 725 KA 200.0-L reaction container initially contains 1.27 kg of N2 and 0.310 kg of H2 at 725 K. Assuming ideal gas behavior, calculate the mass of NH3 (in g) present in the reaction mixture at equilibrium.

FREE Expert Solution

Step 1: Initial Partial Pressure of N2, H2


PV=nRT


molar mass N2214.00 g/mol = 28.00 g/mol

molar mass H221.01 g/mol = 2.02 g/mol


P=nRTVPN2=1.27 kg×103g N21 kg×1 mol N228.00 g N2 0.08206L·atmmol·K725 K200.0 LPN2=(0.310 kg×103g H21 kg×1 mol H22.02 g H2) (0.08206L·atmmol·K)(725 K)(200.0 L)

PN2 = 13.49 atm

PH2 = 45.65 atm


Step 2: ICE Chart


Step 3: Kp expression


Kp=productsreactants=[NH3]2[N2][H2]35.3×10-5=[2x]2[13.49-x][45.65-3x]3

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Problem Details

Ammonia can be synthesized according to the following reaction: N 2(g) + 3H2(g) ⇌ 2NH3(g); Kp = 5.3 x 10–5 at 725 K

A 200.0-L reaction container initially contains 1.27 kg of N2 and 0.310 kg of H2 at 725 K. Assuming ideal gas behavior, calculate the mass of NH3 (in g) present in the reaction mixture at equilibrium.

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