Ammonium chloride, a solid, decomposes to give NH_3(g) and HCl(g). At 345 degree

Question

Ammonium chloride, a solid, decomposes to give NH_3(g) and HCl(g). At 345 degree C, some (s) is placed in an evacuated container. A portion of it decomposes, and the total pressure at equilibrium is 1.21 atm. Extra NH_3(g) is then injected into the container, and when equilibrium is reestablished, the partial pressure of NH_3(g) is 1.30 atm Compute the equilibrium constant in terms of pressures, K_p, for the decomposition of ammonium chloride at 345 degree C. NH_4Cl(s) leftrightarrow NH_3(g) + HCl(g) K = Determine the final partial pressure of HCl(g) in the container. P_HCI = atm

Explanation / Answer

a) According to the equilibrium law, Kp = PNH3 PHCl

ICE table for the decomposition reaction is:

Total pressure is 1.21 atm. The pressure is due to NH3 and HCl. So, x+x = 1.21

2x = 1.21

x = 1.21/2 = 0.605 atm

Kp = PNH3 PHCl = 0.605*0.605 = 0.366

Kp = 0.366

b) After addition of NH3, the equilibrium partial pressure becomes 1.30 atm.

Kp = PNH3 PHCl = 0.366

0.366 = 1.30*PHCl

PHCl = 0.28 atm

PNH4Cl PNH3 PHCl Initial 0 0 0 Change 0 +x +x Equilibirium 0 x x

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