Question

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

Answer 1

Answer: Equilibrium concentration of $B=0.7\times 10^{-3}M$  

Equilibrium concentration of $C=3.17\times 10^{-4}M$  

Explanation:

Initial moles of  $A$ = $1.81\times 10^{-3}$

Volume of container = 1.00 L

Initial concentration of $A=\frac{moles}{volume}=\frac{1.81\times 10^{-3}moles}{1.00L}=1.81\times 10^{-3}M$  

Initial concentration of B=$\frac{moles}{volume}=\frac{1.34\times 10^{-3}moles}{1.00L}=1.34\times 10^{-3}M$  

Initial concentration of $C=\frac{moles}{volume}=\frac{6.37\times 10^{-4}moles}{1.00L}=6.37\times 10^{-4}M$

Equilibrium concentration of $A=2.13\times 10^{-4}M$

The given balanced equilibrium reaction is,

                $A(g)\rightleftharpoons 2B(g)+C(g)$

Initial : $1.81\times 10^{-3}$   $1.34\times 10^{-3}$    $6.37\times 10^{-4}$      

At eqm: $1.81\times 10^{-3}+x$   $1.34\times 10^{-3}-2x$    $6.37\times 10^{-4}-x$    

we are given : $1.81\times 10^{-3}+x=2.13\times 10^{-3}M$

$x=0.32\times 10^{-3}M$

Equilibrium concentration of $B=1.34\times 10^{-3}-2x=1.34\times 10^{-3}-2\times 0.32\times 10^{-3}M=0.7\times 10^{-3}M$  

Equilibrium concentration of $C=6.37\times 10^{-4}-x=6.37\times 10^{-4}-0.32\times 10^{-3}=3.17\times 10^{-4}M$