Be sure to answer all parts. compound a decomposes according to the equation a(g) ⇌ 2 b(g) + c (g) a sealed 1.00−l container initially contains 1.81 × 10−3 mol of a(g), 1.34 × 10−3 mol of b(g), and 6.37 × 10−4 mol of c(g) at 100°c. at equilibrium, [a] is 2.13 × 10−3 m. find [b] and [c]. solve for the equilibrium concentrations of b and c. [b]eq × 10 m [c]eq × 10 m

The equilibrium concentration of B is 1.98  × 10−3 M while the equilibrium concentration of C is 9.57  × 10−4 M.

We have to set up the ICE table as shown below;

Initial concentration of A =       1.81 × 10−3 mol/1.00 L = 1.81 × 10−3 M

Initial concentration of B =   1.34 × 10−3 mol/1. 00 L = 1.34 × 10−3 M

Initial concentration of C =  6.37 × 10−4 mol / 1.00 L = 6.37 × 10−4 M

     A(g) ⇌                      2 B(g)      +            C (g)

I    1.81 × 10−3           1.34 × 10−3                 6.37 × 10−4

C  - x                            +2x                             +x

E 1.81 × 10−3 - x       1.34 × 10−3 + 2x      6.37 × 10−4 + x

We know from the question that the equilibrium concentration of A is  2.13 × 10−3 M so;

2.13 × 10−3  =  1.81 × 10−3 - x

x = 2.13 × 10−3 - 1.81 × 10−3

x = 3.2  × 10−4 M

Hence;

Equilibrium concentration of B

1.34 × 10−3 + 2( 3.2  × 10−4)

= 1.98  × 10−3 M

Equilibrium concentration of C

6.37 × 10−4 + 3.2  × 10−4

= 9.57  × 10−4 M

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Equilibrium concentration of  

Equilibrium concentration of  

Explanation:

Initial moles of   =

Volume of container = 1.00 L

Initial concentration of  

Initial concentration of B=  

Initial concentration of

Equilibrium concentration of

The given balanced equilibrium reaction is,

Initial :            

At eqm:          

we are given :

Equilibrium concentration of  

Equilibrium concentration of  

we can conclude that the melting point of table salt is above room temperature (it's actually 801 deg c). the melting point of mercury must be below room temperature (- 39 deg c).