1) Consider the following reaction at equilibrium. What effect will reducing the

Question

1) Consider the following reaction at equilibrium. What effect will reducing the volume of the reaction mixture have on the system?

Xe(g) + 2 F2(g) ? XeF4(g)

SO2(g) + NO2(g) ? SO3(g) + NO(g) Kc = 0.33

A reaction mixture contains 0.41 M SO2, 0.14 M NO2, 0.12 M SO3 and 0.14 M NO. Which of the following statements is TRUE concerning this system?

a)The reaction will shift in the direction of products

b)The equilibrium constant will decrease.

e)The system is at equilibrium.

Explanation / Answer

1)

3 moles----------> 1 mole (gases)

answer: C

explanation :

Note the number of moles of gas on the left-hand side and the number of moles of gas on the right-hand side. When the volume of the system is changed, the partial pressures of the gases change. If we were to decrease pressure by increasing volume, the equilibrium of the above reaction will shift to the left, because the reactant side has greater number of moles than does the product side. The system tries to counteract the decrease in partial pressure of gas molecules by shifting to the side that exerts greater pressure. Similarly, if we were to increase pressure by decreasing volume, the equilibrium shifts to the right, counteracting the pressure increase by shifting to the side with fewer moles of gas that exert less pressure.

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2)

calculate the K value=

= (0.12*0.14)/(0.41*0.14) =0.292

here K <Kc hence reaction will shift in the direction of products

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