At a certain temperature, a 21.0-l contains holds four gases in equilibrium. their masses are: 3.5 g so3, 4.6 g so2, 13.5 g n2, and 0.98 g n2o. what is the value of the equilibrium constant at this temperature for the reaction of so2 with n2o to form so3 and n2 (balanced with lowest whole-number coefficients) ?

Question

Chemistry

Digger

8 June, 01:04

At a certain temperature, a 21.0-l contains holds four gases in equilibrium. their masses are: 3.5 g so3, 4.6 g so2, 13.5 g n2, and 0.98 g n2o. what is the value of the equilibrium constant at this temperature for the reaction of so2 with n2o to form so3 and n2 (balanced with lowest whole-number coefficients) ?

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Janae Skinner · Answer 1

SO2 + N2OSO3+N2

Kc = concentration of product / concentration of reactant

(SO3) (N2 / (SO2) (N2O)

Find the concentration of each reactant and products

concentration = number of moles / volume in liters

moles of SO3 = 3.5/80=0.044moles

SO2=4.6/64=0.072 moles

N2=13.5/28=0.483 moles

N2O=0.98/44=0.022 moles

concentration is therefore=

SO3=0.044/21 = 2.095 x 10^-3

SO2=0.072 / 21=3.43 x 10^-3

N2=0.483/21=0.023

N2O=0.022/22 = 1.048 x 10^-3

Kc is therefore = { (2.095 x10^-3) (0.023) } / { (3.43 x10^-3) (1.048 x10^-3) }=13.40