For the reaction 2A(g)+B(aq)+3C(l)D(s)+3E(g), the concentrations at equilibrium are found to be

A:2.8×103Pa

B: 1.2×10-2M

C:12.8M

D: 16.5M

E:3.6×104Torr

Find the numerical value of the equilibrium constant that would appear in a conventional table of equilibrium constants.

Short Answer

Expert verified

The numerical value of the equilibrium constant is 1.2×1010.

Step by step solution

01

The equilibrium constant.

Equilibrium constant (K): In the equilibrium reaction, the ratio of the concentration of the reactant and the concentration of the product is taken. If K's value is smaller than 1, the reaction should be moved to the left; if K is more than 1, the reaction should be moved to the right.

Example of an equilibrium reaction:

aA+bBKcC+dD

K=[C]c[D]d[A]a|B|b

02

The numerical value of the equilibrium constant for the given reaction.

The given equation is,

2Ag+Baq+3CIDs+3E(g

A=2.8×103PaB=1.2×10-2MC=12.8MD=16.5ME=3.6×104Torr

The numerical value of the equilibrium constant for the given reaction:

K=PE3PA2B

K=3.6×104Torr760Tor/atm×1.013baratm32.8×103pa105Pabar21.2×10-2M=1.2×1010

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