# Question #0b8eb

May 30, 2015

The equilibrium partial pressure of $H B r$ 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 $H B r$ 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, ${Q}_{p}$

${Q}_{p} = \frac{{\left({H}_{2}\right)}_{0} \cdot {\left({I}_{2}\right)}_{0}}{{\left(H B r\right)}_{0}^{\textcolor{red}{2}}} = \frac{0.010 \cdot 0.010}{{0.20}^{2}} = 0.0025$

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

$\text{ } \textcolor{red}{2} H B {r}_{\left(g\right)} r i g h t \le f t h a r p \infty n s {H}_{2 \left(g\right)} + {I}_{2 \left(g\right)}$
I......0.20...............0.010......0.010
C....(+$\textcolor{red}{2}$x)..................(-x).........(-x)
E....0.20+2x.........0.010-x....0.010-x

The equilibrium constant will be equal to

${K}_{p} = \frac{\left({H}_{2}\right) \cdot \left({I}_{2}\right)}{{\left(H B r\right)}^{\textcolor{red}{2}}} = \frac{\left(0.010 - x\right) \cdot \left(0.010 - x\right)}{0.20 + 2 x} ^ 2$

${K}_{p} = \frac{{\left(0.010 - x\right)}^{2}}{0.20 + 2 x} ^ 2$

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

$\sqrt{\frac{{\left(0.010 - x\right)}^{2}}{0.20 + 2 x} ^ 2} = \sqrt{{K}_{p}}$

$\frac{0.010 - x}{0.20 + 2 x} = \sqrt{4.18 \cdot {10}^{- 9}} = 0.00006465$

This is equivalent to

$0.010 - x = 0.00001293 + 0.0001293 x$

$0.0099871 = 1.0001293 x \implies x = \frac{0.0099871}{1.0001293} = 0.009986$

As a result, the equilibrium partial pressure of $H B r$ will be

${P}_{H B r} = 0.20 + 2 \cdot 0.009986 = \text{0.21997 atm}$

Rounded to two sig figs, the answer will be

${P}_{H B r} = \textcolor{g r e e n}{\text{0.22 atm}}$

Indeed, the equilibrium partial pressure of $H B r$ increased. This equilibrium lies so much to the left, that the equilibrium partial pressures of the two products will be

${P}_{{H}_{2}} = {P}_{B {r}_{2}} = 0.010 - 0.009986 = \text{0.000014 atm}$

May 30, 2015

The concentration = $3.84 \text{mol""/} {m}^{3}$

$2 H B {r}_{\left(g\right)} r i g h t \le f t h a r p \infty n s {H}_{2 \left(g\right)} + {I}_{2 \left(g\right)}$

Initial partial pressures (Atm):

$H B r = 0.2$

${H}_{2} = 0.01$

$B {r}_{2} = 0.01$

At equilibrium we can assume x mol of ${H}_{2}$ and x mol $B {r}_{2}$ are used up.

This means that the partial pressure of $H B r$ 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 $H I$ = 0.2 + (2 x 0.01) = 0.22 Atm

$P V = n R T$

Concentration $\frac{n}{V} = \frac{P}{R T}$

$= \frac{0.22 \times 1.0132 \times {10}^{5}}{8.31 \times 698}$

$= 3.84 \text{mol""/} {m}^{3}$