AP Chemistry Question (Equilibrium)

Test Preparation AP Test Preparation
1

<p>Doesn't understand how to solve (c)& its answer!!!</p>

<p>Just in case, I will write a whole question.</p>

<ol>
<li>At elevated tmeperatures, SbCl5 gas decomposes into SbCl3 gas and Cl2 gas as shown by the following equation:</li>
</ol>

<p>SbCl5(g) -> SbCl3 (g) + Cl2 (g)</p>

<p>a) An 89.7-gram sample of SbCl5 (molecular weight 299.0) is placed in an evacuated 15.0-liter container at 182 °C.
(1) What is concentration in moles/liter of SbCl5 before any decomposition occurs?
(2) What is the pressure in atm of SbCl5 in the container before any decomposition occurs?</p>

<p>b) If the SbCl5 is 29.2 percent decomposes when equilibrium is established at 182 °C, calculate the value for either equilibrium constant, Kp or Kc, for this decomposition reaction. </p>

<p>c) In order to produce some SbCl5, a 1.00-mole sample of SbCl3 is first placed in an empty 2.00-liter container maintained at a temperature different from 182 °C. At this temperature Kc equals 0.117. How many moles of Cl2 must be added to this container to reduce the number of moles of SbCl3 to 0.700 mole at equilibrium?</p>

<p>*Answer to (c)</p>

<p>K = ([SbCl3][Cl2]) ¯ [SbCl5] = 0.117
Equilibrium concentrations:
[SbCl5] = (1.00 - 0.70) mol / 2.00 L = 0.15 M <<---why is it 1.00-0.70???
[SbCl3] = 0.700 mole / 2.00L = 0.350 M
[Cl2] = x
Kc = [(0.350) (x)] ÷ (0.15) = 0.117
x = [Cl2] = 0.50 M
Moles Cl2 at equilibrium = 0.050 mol L x 2.00 L = 0.10 mol
Moles Cl2 needed to make 0.300 mol SbCl3 into SbCl5 = 0.30 mol
Moles Cl2 that must be added = 0.40 mol</p>

<p>Thanks XD</p>

2

<p>Since 0.3 moles of the SbCL3 are becoming SbCl5, that is why you have 1-0.7</p>

3

<p>Ignore the <strong><em>'s in my post. They’re just there to make spaces (when I posted this, all the spaces disappeared, so I replaced them with _</em></strong>'s.)</p>

<p>Do you know how to do an ICE table? Basically they’re doing the same thing, but without the table. If you’ve never learned how to do an ICE table, you should ask your teacher about it.</p>

<p><strong>SbCl5(g) -></strong>SbCl3 (g)<strong>+</strong><em>Cl2 (g)
I</em><em>0 mol</em><strong><em>1.00 mol</em></strong><em>n (I’m using n as the variable for Cl2.)
C</em><em>+x</em><strong><em>-x</em></strong><strong><em>-x
E</em></strong>x__<strong><em>-1.00 mol-x</em></strong>___n-x</p>

<p>For this part of the problem, Cl2 doesn’t matter. The question says that SbCl5 should have 0.700 mol at equilibrium. That means x = 0.700 mol. If you plug that into the x for SbCl3 at equilibrium (1.00 mol-x), you get 1.00 mol-0.700 mol.</p>