Answer:
1.65 L
Explanation:
The equation for the reaction is given as:
A + B ⇄ C
where;
numbers of moles = 0.386 mol C (g)
Volume = 7.29 L
Molar concentration of C = [tex]\frac{0.386}{7.29}[/tex]
= 0.053 M
A + B ⇄ C
Initial 0 0 0.530
Change +x +x - x
Equilibrium x x (0.0530 - x)
[tex]K = \frac{[C]}{[A][B]}[/tex]
where
K is given as ; 78.2 atm-1.
So, we have:
[tex]78.2=\frac{[0.0530-x]}{[x][x]}[/tex]
[tex]78.2= \frac{(0.0530-x)}{(x^2)}[/tex]
[tex]78.2x^2= 0.0530-x[/tex]
[tex]78.2x^2+x-0.0530=0[/tex]
Using quadratic formula;
[tex]\frac{-b+/-\sqrt{b^2-4ac} }{2a}[/tex]
where; a = 78.2 ; b = 1 ; c= - 0.0530
= [tex]\frac{-b+\sqrt{b^2-4ac} }{2a}[/tex] or [tex]\frac{-b-\sqrt{b^2-4ac} }{2a}[/tex]
= [tex]\frac{-(1)+\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}[/tex] or [tex]\frac{-(1)-\sqrt{(1)^2-4(78.2)(-0.0530)} }{2(78.2)}[/tex]
= 0.0204 or -0.0332
Going by the positive value; we have:
x = 0.0204
[A] = 0.0204
[B] = 0.0204
[C] = 0.0530 - x
= 0.0530 - 0.0204
= 0.0326
Total number of moles at equilibrium = 0.0204 + 0.0204 + 0.0326
= 0.0734
Finally, we can calculate the volume of the cylinder at equilibrium using the ideal gas; PV =nRT
if we make V the subject of the formula; we have:
[tex]V = \frac{nRT}{P}[/tex]
where;
P (pressure) = 1 atm
n (number of moles) = 0.0734 mole
R (rate constant) = 0.0821 L-atm/mol-K
T = 273.15 K (fixed constant temperature )
V (volume) = ???
[tex]V=\frac{(0.0734*0.0821*273.15)}{(1.00)}[/tex]
V = 1.64604
V ≅ 1.65 L
