What are the equilibrium partial pressures of \ m PCl_3 , \ m Cl_2, and \ m PCl_

Question

What are the equilibrium partial pressures of m PCl_3 , m Cl_2, and m PCl_5, respectively? For the exothermic reaction PCl3(g)+Cl2(g) ightleftharpoons PCl_5(g) K_{ m p} = 0.100 at a certain temperature. A flask is charged with 0.500 m atm m PCl_3 , 0.500 m atm m Cl_2, and 0.300atm m PCl_5 at this temperature.

Explanation / Answer

PCl3 + Cl2 ---> PCl5 K=[PCl5]/([PCl3][Cl2] =0.3/(0.5x0.5) = 1.2 since K sholuld be equal to 0.1 so the reaction will be in the reverse direction let the PCl3 decomposed =x atm so, K = 0.1= (0.3-x)/{(0.5+x)(0.5+x)} x=-11.23 or 0.244 so we select x = 0.244 equilibrium partial pressure of PCl3=0.3-0.244=0.056atm equilibrium partial pressure of Cl2=0.5+0.244=0.744 atm equilibrium partial pressure of PCl3=0.5+0.244=0.744 atm

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