3) Consider the reaction: 2S02(g) + O2(g) rightleftharpoons 2SO3(g) Delta H? = -

Question

3) Consider the reaction: 2S02(g) + O2(g) rightleftharpoons 2SO3(g) Delta H? = - 196.6 kJ/mol The equilibrium is displaced to the left if: A) the temperature is raised B) some sulfur dioxide is added C) the temperature is lowered D) the pressure is raised E) some sulfur trioxide is removed 4) For the reaction; N2(g) + 3 H2(g) rightleftharpoons 2 NH3(g), the equilibrium amount of NH3 will be increased 4) by: I) increasing the pressure II) adding H2 III) removing N2 IV) decreasing the pressure A) 1, III B) I, II C) II, 111 D) III only E) II, iV 5) For the reaction: H2(g) + 12(g) 2 HI(g), Kc 92.0 5) When equilibrium concentrations of HI and 12 are [HI] = 0.115 M and [12] = 0.250 M, the equilibrium concentration of [H2] is: A) 5.75 X 10-4 M B) 1.74 x i3 M (show work for pts) C) 9.56 M D) 0.135 M E) 5.00 x in-3 M 6) For the reaction 2 S02 (g) + 02 (g) 2 S03 (g), Kc 2.8 x 102 at 1000 K. If a vessel is filled with 6) these gases such that the initial concentrations are [S021 = 0.025, [021 = 0.035, and [S03] = 0.046, in which direction will a reaction occur and why? A) toward products because Q = 96 B) toward reactants because Q = 2.8 x io3 C) toward reactants because Q = 0.019 (show work for pts) D) toward products because Q = 53 E) it is at equilibrium because Q = I 7) Consider the following reaction at a certain temperature. 7) 2S03(g) 2SO(g) + 02(g) When the initial concentration of S03(g) is 0.128 M, the concentration of oxygen gas at equilibrium is found to be 0.0130 M. Calculate K for this reaction. A) 8.45x i04 B) I.62x i02 C) l.47x i03 D) 7.64x i05 (show work for pts)

Explanation / Answer

3] A} when the tmp is increased the reaction moves to the left as it start sconsuming the heat

4] A] as increasing the pressure and removing N2 both favours the reaction in the forward direction

5] Kc= [HI]^2 / [H2][I2]

92 = [0.115]^2/[H2][0.25]

[H2] = 1.74*10^3M

6]Q = [SO3]^2 / [SO2][O2]

= 96.73

as Q< K

so the answer is A , towards product

7 ]

2So3 ---> 2SO2 + O2

0.128 0 0

at euili brium

= 0.128 - 2x 2x x

x = 0.0130

so

0.128-2*0.0130 = SO3

2* 0.0130 = SO2

O2= 0.0130

so Kc = 0.0130* 0.026^2 / 0.102^2

= kc = 8.44 * 10^-4

A

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