Note: quadratic formula for ax^2 + bx + c = 0 has solution: x = -b plus-minus(b^

Question

Note: quadratic formula for ax^2 + bx + c = 0 has solution: x = -b plus-minus(b^2 - 4ac)^1/2/2a Calculate the numerical value of K_p for the following reaction, given the information below: 1/4 S_g(s) + 3 O_2(g) equivalent to 2 SO_3(g) K_p = ? Know: 1/8 S_g(s) + O_2(g) equivalent to SO_2(g) K_p = 4.0 times 10^52 SO_3(g) equivalent to SO_2(g) + 1/2 O_2(g) K_p = 3.6 times 10^-13 Which direction would the reaction shift (left, right, no change) if you made the following changes to the reaction at equilibrium? Rxn: 2 CO(g) + O_2(g) equivalent to 2 CO_2(g) Rxn is exothermic Add O_2 gas decrease temperature of reaction move the reaction to a smaller volume add inert He gas

Explanation / Answer

Q.1: Multiplying the first given equation by 2 we get

1/4 S8(s) + 2O2(g) -----> 2SO2(g); Kp1 = (4.0x1052)2 = 1.6x10105 -------- (1)

Multiplying the second given equation by 2 and then reversing we get

2SO2(g) + O2(g) ---- > 2SO3(g); Kp2 = 1 / (3.6x10-13)2 = 7.716x1024 ------ (2)

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By adding eqn(1) and (2) we get

1/4 S8(s) + 3O2(g) -----> 2SO3(g); Kp = Kp1 x Kp2 =  1.6x10105 x 7.716x1024 = 1.235x10130 (answer)

Q:2: (a) When we add O2 gas it will react with more and more CO(g) to produce CO2(g). Hence the equilibrium will shift towards right.

(b) Since the reaction is exothermic, decreasing the temerature will shift the reaction towards right.

(c) Decreasing the volume increases the pressure. When pressure increases the reaction moves in a direction where number of gaseous moles decreases. Since the product contains less number of moles, the equilibrium shifts towards right.

(d) When we add incert gas there is no chage in the equilibrium constant. Hence there would be no change in the reaction.

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