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  

2b2t

explanation:

2b2t

that's what i got you can count the number of atoms