Hey i need some help on this particular problem, i will award 1500 points to who
ID: 819626 • Letter: H
Question
Hey i need some help on this particular problem, i will award 1500 points to whoever gets it right using sig figs and have units on all
Kc = 4.00 at 25 degree C for the reaction 2J(g) Q(g). To an initially empty 10.0 L container maintained at 25 degree C were added 0.500 mol of gaseous J and 1.000 mol of gaseous Q. Assume J and Q behave ideally. When equilibrium was attained in the container, was there greater than 0.750 mol Q, less than 0.750 mol Q, 0.750 mol Q, or cannot be determined present in the container? Justify your choice. When equilibrium was attained in the container, was the total pressure inside the container greater than 3.67 atm, less than 3.67 atm, equal to 3.67 atm, or cannot be determined? Justify your choice.Explanation / Answer
17) initial moles of J = 0.5
initial moles of Q = 1
at equilibrium J moles = 0.5-2x
at equi Q moles = 1+x
now K = [Q]/[J]^2
4 =[ ( 1+x) /10 ] / [( 0.5-2x) /10 ]^2
4 = ( 1+x)10/( 0.5-2x) ^2
4( 4x^2 -2x + 0.25) = 10+10x
16x^2 -18x -9 = 0
x = 1.5 or -0.375
we take -0.375 value ( as 1.5 is higher amount than present)
hence Q at equi = 1-0.375 = 0.625
hence Q amount is less than 0.75 mol
b) J moles at equi = 0.5- 2(-0.375) = 0.5+2(0.375)
total moles at equi J moles + Q moles = (0.5+2(0.375) + 0.625 = 1.25+0.625 = 1.875 =n
PV = nRT is eq , R = 0.0821 liter atmK^-1 , V = 10 L , T = 25C = 25+273 = 298 K ,
P = nRT/V = ( 0.0821 x 298 x 1.875 /10)
= 4.59 atm
hence Pressure is greater than 3.67 atm