Self Test 5I.3b

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gabbymaraziti
Posts: 111
Joined: Wed Sep 18, 2019 12:19 am

Self Test 5I.3b

Postby gabbymaraziti » Wed Jan 15, 2020 9:10 pm

I'm not sure how to start the calculations for Self Test 5I.3b. Can anyone explain how to begin?

Hydrogen chloride gas is added to a reaction vessel containing solid iodine until its partial pressure reaches 0.012 bar. At the temperature of the experiment, K = 3.5 x 10^-32 for 2HCl(g) + I2(s) -> 2HI(g) + Cl2(g). Assume that some I2 remains at equilibrium. What are the equilibrium partial pressures of each gaseous substance in the reaction mixture?

I know that I2 is in its solid form, so it is not included in the equilibrium constant, but I'm still not sure how to begin with the two values we're given.

Tyler Angtuaco 1G
Posts: 130
Joined: Wed Sep 11, 2019 12:16 am

Re: Self Test 5I.3b

Postby Tyler Angtuaco 1G » Wed Jan 15, 2020 9:28 pm

I would assume that the pressure given is the initial pressure for HCl since the question does not explicitly state it is an equilibrium value, and also because the question seems to phrase it as if it just added HCl until there was enough for the desired reaction to occur. Next, I would construct an ICE table since you know the initial values of HCl (which was given) and the products (0). Since K is given, you can calculate the equilibrium partial pressures using the equilibrium expressions from the ICE table. Keep in mind the K they provide is very small (less than 10^-4), so it may help to ignore the "change" in initial concentration (denoted by x) in your equilibrium expression for HCl.

805303639
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Re: Self Test 5I.3b

Postby 805303639 » Wed Jan 15, 2020 9:39 pm

Hi Gabby. Yes, I2 is not included in the equilibrium expression K= (PCl2)(PHI)^2/(PHCl)^2. Since we know the K value, the initial partial pressure of HCl, and the stoichiometric coefficients, we can use an ICE table to solve for the equilibrium partial pressures. Set up an ice table with the initial value for HCl as 0.012 and 0 for HI and Cl2. Use the coefficients to determine the changes in partial pressures. Plugging into the K expression, K = 3.5 x 10^-32 = (2x)^2 (x)/(0.012-2x)^2. Since 10^-34 < 10^-3, we can treat the denominator as essentially unchanged, 0.012^2. Rearrange and solve for x.

Skyllar Kuppinger 1F
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Joined: Thu Jul 25, 2019 12:16 am

Re: Self Test 5I.3b

Postby Skyllar Kuppinger 1F » Sat Jan 18, 2020 6:08 pm

set up an ice table. it seems that 0.012 bar is simply the initial value for HCl (but they did word it kinda confusingly, at first I thought they were trying to say that 0.012 is the equilibrium value). the initial values of HI and Cl2 will be 0. ignore I2 bc it is a solid. Under the C row in the ice table, it will be -2x for HCl, ignore I2, +2x for HI, and +x for Cl2. You then need to make your equation by raising the equilibrium concentrations to their appropriate powers. However, you can get rid of the x in the denominator bc K is VERY VERY small. then simply solve for x and use that x to find the partial pressures of each gas.

Skyllar Kuppinger 1F
Posts: 52
Joined: Thu Jul 25, 2019 12:16 am

Re: Self Test 5I.3b

Postby Skyllar Kuppinger 1F » Sat Jan 18, 2020 6:10 pm

correction-- in my previous reply I meant to say equilibrium partial pressures, not equilibrium concentrations.


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