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Solved quiz for Chemical Principles course in class of Dr. Brian Lamp at Truman State University
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CHEM 130 Name _____________________________________________ Quiz 9 – November 17, 2017
Complete the following problems. Write your final answers in the blanks provided.
There are several ways to approach this. In the end, we need equilibrium concentrations of all species.
2SO 3 2SO 2 + O 2 I 0.760 mol/5 L = 0.152 M 0 0 C -2x +2x +x E 0.152-2(0.032) =0.088 M 0+2(0.032) =0.064 M 0.160 mol/5L = 0. 0 320 M =x
Kc = [SO 2 ]^2 [O 2 ] = (0.0640)^2 (0.0320) = 0. [SO 3 ]^2 (0.088)^2
Answer_______ 0.0169 ________________
We need to rearrange reactions to make our target reaction:
2(NO 2 (g) ½ N 2 (g) + O 2 (g)) Kc = 1/(1.0 x 10-^9 )^2 2(NOCl(g) + ½ O 2 (g) NO 2 Cl(g)) Kc = (1.1 x 10^2 )^2 2(NO 2 Cl(g) NO 2 (g) + ½ Cl 2 (g)) Kc = 1/(0.3)^2
So, the sum of the reactions is:
2NO 2 + 2NOCl + O 2 + 2NO 2 Cl N 2 + 2O 2 + 2NO 2 Cl + 2NO 2 + Cl 2 2NO 2 Cl N 2 + O 2 + Cl 2
Kc = (1.1 x 10^2 )^2 = 1.3 x10^23 (1.0 x 10-^9 )^2 (0.3)^2
Answer______ 1.3 x10^23 _________________
PCl 5 PCl 3 + Cl 2 I 0.280 mol/4 L = 0.0700 M 0 0 C -x +x +x E 0.0700-x x x
Kc = [PCl 3 ][Cl 2 ] = (x)(x) [PCl 5 ] 0.0700-x
Kc = x^2 0.0700-x
(0.0700-x)Kc = x^2 0.0700Kc-xKc = x^2 x^2 + Kcx – 0.0700Kc = 0 Inserting 1.80 for Kc x^2 + 1.80x – 0.126 = 0 Using the quadratic equation with a = 1, b = 1.80 and c = -0.126 gives: x = 0.067 or -1. Since we’ve defined x as a concentration, a negative value makes no chemical sense, therefore the appropriate value for x is 0.067 or [PCl 3 ] = [Cl 2 ] = 0.0675 M and [PCl 5 ] = 0.0700-0.0675 = 0.0025 M
Answer_ [PCl 3 ] 0.0675 M and [PCl 5 ] = 0.0025 M_
Possibly Useful Information
2 1
2 1 x x
y y x
y slope m −
2 a
b b 4 ac x
= R = 0.08206 L atm mol
R = 8.314 J mol-1^ K-
pV = nRT ∆G = -RTlnK Kp = Kc(RT)∆n