Question #0b8eb

The equilibrium partial pressure of #HBr# will be equal to 0.22 M.

Once again, before doing any actual calculations, take a second to try and predict what's going to happen when you mix those amounts of the gases together.

Notice that the equilibrium constant, #K_p#, is much smaller than 1, which implies that this equilibrium will favor the reactant almost exclusively.

You could predict that the equilibrium partial pressure of #HBr# will actually be larger than the initial value of 0.20 atm, which implies that the reverse reaction will proceed until equilibrium is reached.

To confirm this, calculate the reaction quotient, #Q_p#

#Q_p = ((H_2)_0 * (I_2)_0)/((HBr)_0^color(red)(2)) = (0.010 * 0.010)/(0.20^2) = 0.0025#

Since #Q_p# is signifcantly larger than #K_p#, the equilibrium will indeed shift to the left, favoring the formation of even more reactant. Use an ICE table to help you with the calculations

#" "color(red)(2)HBr_((g)) rightleftharpoons H_(2(g)) + I_(2(g))#
I......0.20...............0.010......0.010
C....(+#color(red)(2)#x)..................(-x).........(-x)
E....0.20+2x.........0.010-x....0.010-x

The equilibrium constant will be equal to

#K_p = ((H_2) * (I_2))/((HBr)^color(red)(2)) = ((0.010-x) * (0.010-x))/(0.20+2x)^2#

#K_p = ((0.010-x)^2)/(0.20+2x)^2#

Take the square root of both sides of the equation to get

#sqrt(((0.010-x)^2)/(0.20+2x)^2) = sqrt(K_p)#

#(0.010-x)/(0.20 + 2x) = sqrt(4.18 * 10^(-9)) = 0.00006465#

This is equivalent to

#0.010 - x = 0.00001293 + 0.0001293x#

#0.0099871 = 1.0001293x => x = 0.0099871/1.0001293 = 0.009986#

As a result, the equilibrium partial pressure of #HBr# will be

#P_(HBr) = 0.20 + 2 * 0.009986 = "0.21997 atm"#

Rounded to two sig figs, the answer will be

#P_(HBr) = color(green)("0.22 atm")#

Indeed, the equilibrium partial pressure of #HBr# increased. This equilibrium lies so much to the left, that the equilibrium partial pressures of the two products will be

#P_(H_2) = P_(Br_2) = 0.010 - 0.009986 = "0.000014 atm"#

The concentration = # 3.84 "mol""/"m^(3)#

#2HBr_((g))rightleftharpoonsH_(2(g))+I_(2(g))#

Initial partial pressures (Atm):

#HBr = 0.2#

#H_2=0.01#

#Br_2=0.01#

At equilibrium we can assume x mol of #H_2# and x mol #Br_2# are used up.

This means that the partial pressure of #HBr# is (0.2 +2x)

Because the value of Kp is so small I am going to assume that the reaction virtually goes to completion right to left.

This means that the equilibrium partial pressure of #HI# = 0.2 + (2 x 0.01) = 0.22 Atm

#PV=nRT#

Concentration #n/V=(P)/(RT)#

#=(0.22xx1.0132xx10^(5))/(8.31xx698)#

#=3.84"mol""/"m^(3)#