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Question #0b8eb

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Answer 1 · 2015-05-30T17:44:45

The equilibrium partial pressure of $H B r$ $HBr$ 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, $K_{p}$ $K_p$ , 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 $H B r$ $HBr$ 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, $Q_{p}$ $Q_p$

$Q_{p} = \frac{{(H_{2})}_{0} \cdot {(I_{2})}_{0}}{{(H B r)}_{0}^{2}} = \frac{0.010 \cdot 0.010}{{0.20}^{2}} = 0.0025$ $Q_p = ((H_2)_0 * (I_2)_0)/((HBr)_0^color(red)(2)) = (0.010 * 0.010)/(0.20^2) = 0.0025$

Since $Q_{p}$ $Q_p$ is signifcantly larger than $K_{p}$ $K_p$ , 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

$2 H B r_{(g)} ⇌ H_{2 (g)} + I_{2 (g)}$ $" "color(red)(2)HBr_((g)) rightleftharpoons H_(2(g)) + I_(2(g))$
I......0.20...............0.010......0.010
C....(+ $2$ $color(red)(2)$ x)..................(-x).........(-x)
E....0.20+2x.........0.010-x....0.010-x

The equilibrium constant will be equal to

$K_{p} = \frac{(H_{2}) \cdot (I_{2})}{{(H B r)}^{2}} = \frac{(0.010 - x) \cdot (0.010 - x)}{{(0.20 + 2 x)}^{2}}$ $K_p = ((H_2) * (I_2))/((HBr)^color(red)(2)) = ((0.010-x) * (0.010-x))/(0.20+2x)^2$

$K_{p} = \frac{{(0.010 - x)}^{2}}{{(0.20 + 2 x)}^{2}}$ $K_p = ((0.010-x)^2)/(0.20+2x)^2$

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

$\sqrt{\frac{{(0.010 - x)}^{2}}{{(0.20 + 2 x)}^{2}}} = \sqrt{K_{p}}$ $sqrt(((0.010-x)^2)/(0.20+2x)^2) = sqrt(K_p)$

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

This is equivalent to

$0.010 - x = 0.00001293 + 0.0001293 x$ $0.010 - x = 0.00001293 + 0.0001293x$

$0.0099871 = 1.0001293 x \Rightarrow x = \frac{0.0099871}{1.0001293} = 0.009986$ $0.0099871 = 1.0001293x => x = 0.0099871/1.0001293 = 0.009986$

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

$P_{H B r} = 0.20 + 2 \cdot 0.009986 = 0.21997 atm$ $P_(HBr) = 0.20 + 2 * 0.009986 = "0.21997 atm"$

Rounded to two sig figs, the answer will be

$P_{H B r} = 0.22 atm$ $P_(HBr) = color(green)("0.22 atm")$

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

$P_{H_{2}} = P_{B r_{2}} = 0.010 - 0.009986 = 0.000014 atm$ $P_(H_2) = P_(Br_2) = 0.010 - 0.009986 = "0.000014 atm"$

Answer 2 · 2015-05-30T18:43:01

The concentration = $3.84 mol / m^{3}$ $3.84 "mol""/"m^(3)$

$2 H B r_{(g)} ⇌ H_{2 (g)} + I_{2 (g)}$ $2HBr_((g))rightleftharpoonsH_(2(g))+I_(2(g))$

Initial partial pressures (Atm):

$H B r = 0.2$ $HBr = 0.2$

$H_{2} = 0.01$ $H_2=0.01$

$B r_{2} = 0.01$ $Br_2=0.01$

At equilibrium we can assume x mol of $H_{2}$ $H_2$ and x mol $B r_{2}$ $Br_2$ are used up.

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

$P V = n R T$ $PV=nRT$

Concentration $\frac{n}{V} = \frac{P}{R T}$ $n/V=(P)/(RT)$

$= \frac{0.22 \times 1.0132 \times 10^{5}}{8.31 \times 698}$ $=(0.22xx1.0132xx10^(5))/(8.31xx698)$

$= 3.84 mol / m^{3}$ $=3.84"mol""/"m^(3)$

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