HW701AN

CHEMISTRY 113	HOMEWORK 7	NAME____________________
Due Feb 28, 2001		Comp.Homework: 17, Equil: 1-5

1.. For the reaction NH₃(g) + CH₃COOH(l) = NH₂CH₂COOH(s) + H₂(g), /\H^o = -6.90 kJ/mol. Predict the effect of the following changes on the system at equilibrium.

a) Increase in temperature Rxn is exothermic; toward reactants to absorb energy	Shift left	Shift right	Unchanged
b) Decrease volume of container increases pressure; to lower pressure, fewer moles of gas	Shift left	Shift right	Unchanged
c) Add catalyst does not affect equilibrium; only speeds up reaching equilibrium	Shift left	Shift right	Unchanged
d) Remove H₂(g) equilibrium will shift to provide more H₂(g)	Shift left	Shift right	Unchanged
e) Add CH₃COOH(l) concentration of a pure liquid is unchanged as long as some is present	Shift left	Shift right	Unchanged

2. A gaseous reaction mixture at 460^oC contains 0.00250 mol SO₂, 0.00350 mol NO₂, 0.0250 mol NO and 0.0400 mol SO₃in a 2.0 L container; Kc = 85.0 for SO₂(g) + NO₂ (g) = SO₃ (g) + NO(g).

a) Is the system at equilibrium? NO YES Explain.

K_c = {[SO₃][NO]} / {[SO₂][NO₂]

à Q = {[0.0400mol/2.0L][0.0250 mol/2.0 L]} / {[0.00250 mol/2.0L][0.00350mol/2.0L]} = 114.3

Q= 114.3 =/= K_c = 85.0 so that the system is not at equilibrium

b) If it is not at equilibrium, in which direction will the system move to reach equilibrium? SHOW REASONING.

Since Q is larger than K_c, there are too many products so that the equilibrium will shift toward the reactants (left)

3. The value of Kp or Kc is given for each of the following reactions. Write the Kp expression for each reaction and calculate the value of the one of Kp or Kc, which is not given.

a) 2 NOBr(g) = 2 NO(g) + Br₂(g); Kc = 0.1563 at 350 K

K_p =( pNO ² * pBr₂ ) / pNOBr ²

K_p = K_c(RT)^Dng = 0.1563 (0.08206 Latm/mol deg * 350K) ^(3-2) = 0.1563(28.72)¹ = 4.489

Kp = |_4.49 (atm)____

b) CO(g) + 2 H₂(g) = CH₃OH(g); Kp = 2.05 x 10⁴ at 298 K

K_p = pCH₃OH / (pCO * pH₂ ²)

K_c = K_p(RT) ^-Dng = 2.05 x 10⁴ (0.08206 Latm/mol deg * 298K)^(1-3) = 2.05x10⁴ (24.45)² = 1.226 x 10⁷

Kc = |_1.23 x 10⁷ (L²/mol²)___

4. When 0.0560 mol O₂ and 0.20 mol N₂O are put in a 1.00 liter sealed container at 25^oC, the following equilibrium is established: 2N₂O(g) + 3 O₂(g) = 4NO₂(g). The equilibrium concentration of NO₂ is 0.020 mol/L. Calculate:

K_c = [NO₂]⁴ / {[N₂O]²[O₂]³}

	N₂O	O₂	NO₂
start	0.20 M	0.0560 M	none
change	-2x	-3x	+4x
At equilibrium	0.20 - 2x	0.0560 -3x	4x = 0.020 M

X = 0.0050 M

a) [N₂O]

[N₂O] = |_0.20 - 0.010 M = 0.19 M_______

b) [O₂]

[O₂] = |_0.0560 M - 0.015 M = 0.041 M________

c) Kc =[NO₂]⁴ / {[N₂O]²[O₂]³} = [0.020]⁴ / {[0.19]²[0.041]³} = 1.6 x 10⁷ / 2.49 x 10^-6 = 0.06431

Kc = |_0.064______

5. For the reaction 2 IBr(g) = I₂(g) + Br₂(g), Kc is 2.5 x 10^-3 at 25^o C. Calculate the equilibrium concentration of each species in a 2.0 L vessel starting with 0.30 mol IBr.

K_c = {[I₂][Br₂]} / [IBr]²

	IBr	I₂	Br₂
Start	0.30 mol/2.0 L = .15M	none	None
Change	- 2x	+ x	+ x
At equilibrium	0.15 - 2x	x	x

K_c = {[I₂][Br₂]} / [IBr]² = [x][x] / [0.15 - 2x] = 2.5 x 10^-3;

solve by taking square root of both sides of equation

x / (0.15 - 2x) = 5.0 x 10^-2; x = (0.15 - 2x) * 5.0 x 10^-2 = 7.5 x 10^-3 - 0.10 x

1.10 x = 7.5 x 10^-3 ; x = 6.8182 x 10^-3

[IBr] = |_0.15 - 2(.00682) = 0.1364 = 0.14 M_____

[I₂] = |_x = 0.0068 M______

[Br₂] = |_= x = 0.0068 M_______________

6. The Kp (atm) = 0.108 for the decomposition of ammonium hydrogen sulfide according to the reaction: NH₄HS(s) = NH₃(g) + H₂S(g). Calculate the equilibrium partial pressures of all species

K_p = 0.108 = p NH₃ * p H₂S = x * x = x² since the ammonium bisulfide is a solid

x = 0.3286 atm

p NH₃ = |_0.329 atm________

p H₂S = |_0.329 atm ________

7. The heterogeneous reaction 2 HCl(g) + I₂(s) = 2 HI(g) + Cl₂(g) has Kc = 1.6 x 10^-34 at 25^oC. If 1.00 mol of HCl and solid I₂ was placed in a 1.00 liter container, what would be the equilibrium concentrations of HI and Cl₂ in the container?

K_c = {[HI]²[Cl₂]} / [HCl]²; as long as there is some left, pure I₂ solid is not part of the K_c expression

	HCl	HI	Cl₂
Start	1.00mol/1.00 L = 1.00 M	none	none
Change	-2x	+2x	+ x
At equilibrium	1.00 - 2x	2x	x

K_c = {[HI]²[Cl₂]} / [HCl]² = {[2x]²[x] / [1.00 - 2x]² = 1.6 x 10^-34; K_c is very small so approximate 2x as very small compared to 1.00 M;

4x³/ [1.00]² ~ 1.6 x 10^-34 ~ 4 x³;

x³ ~ 1.6 x 10^-34/4 ~ 4.0 x 10^-35

x ~ 3.420 x 10^-12 M ; very small approximation OK

[HI] = |_ 2(3.42 x 10^-12)M = 6.84 x 10^-12 M_________

[Cl₂] = |_3.42 x 10^-12 M _______________

8.a. Give the conjugate acid for each of the following. (Add H⁺)

N₂H₄ _N₂H₅⁺______ H₂PO₄^- _H₃PO₄_____ SO₄^2- _HSO₄^- _____ HF _H₂F⁺_____

b. Give the conjugate base for each of the following. (Subtract H⁺)

NH₃_NH₂^-_____ HPO₄^2- _PO₄^3- _____H₂SO₃ _HSO₃^-____ HCl _Cl ^-____

9. Choose the stronger acid. State the reason.

HOCl or HClO₂

The extra very electronegative O (not bonded to H) helps withdraw electron density making it easier to remove H⁺

HBr or HCl

The weaker bond and less electronegative Br makes it easier to remove H⁺

H₂CO₃ or HNO₃

The extra very electronegative O (not bonded to H), together with the more electronegative N helps withdraw electron density making it easier to remove H⁺