Question #b584a

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Question #b584a

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Answer 1 · 2017-06-05T18:26:35

$H_{2}$ $"H"_2$ : $0.322$ $0.322$ $M$ $M$
$I_{2}$ $"I"_2$ : $0.00749$ $0.00749$ $M$ $M$
$HI$ $"HI"$ : $0.347$ $0.347$ $M$ $M$

Explanation:

We're asked to find the molar concentrations of the substances at equilibrium from the given data.

From the chemical equation

$H_{2} (g) + I_{2} (g) ⇌ 2 HI (g)$ $"H"_2 (g) + "I"_2 (g) rightleftharpoons 2"HI" (g)$

the equilibrium-constant expression is

$K_{c} = \frac{{[HI]}^{2}}{[H_{2}] [I_{2}]}$ $K_c = (["HI"]^2)/(["H"_2]["I"_2])$

The given volume is $1$ $1$ ${dm}^{3}$ $"dm"^3$ , which is equivalent to $1$ $1$ $L$ $"L"$ , so the given mole values in the equation are the same as their molar concentrations.

Let's create a makeshift I.C.E. chart, using bullet points:

Initial Concentration ( $M$ $M$ ):

$H_{2}$ $"H"_2$ : $0.496$ $0.496$
$I_{2}$ $"I"_2$ : $0.181$ $0.181$
$HI$ $"HI"$ : $0$ $0$

Change in Concentration ( $M$ $M$ ):

$H_{2}$ $"H"_2$ : $- x$ $-x$
$I_{2}$ $"I"_2$ : $- x$ $-x$
$HI$ $"HI"$ : $+ 2 x$ $+2x$

Final Concentration ( $M$ $M$ ):

$H_{2}$ $"H"_2$ : $0.496 - x$ $0.496-x$
$I_{2}$ $"I"_2$ : $0.00749$ $0.00749$
$HI$ $"HI"$ : $2 x$ $2x$

Since we're given both the initial and final concentrations for $I_{2}$ $"I"_2$ , we can calculate the change in concentration, and thus $x$ $x$ :

$x = overbrace(0.181)^("initial I"_2)-overbrace(0.00749)^("final I"_2) = color(red)(0.17351$

Now that we know $x$ , we can use this to calculate the remaining equilibrium concentrations:

Final Concentrations ( $M$ ):

$"H"_2$ : $0.496-color(red)(0.17351) = 0.322$
$"I"_2$ : $0.00749$
$"HI"$ : $2(color(red)(0.17351)) = 0.347$

The equilibrium concentrations are thus

$"H"_2$ : $0.322$ $M$
$"I"_2$ : $0.00749$ $M$
$"HI"$ : $0.347$ $M$

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Question #b584a

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