20) Consider the following reaction at equilibrium. What effect will decreasing

Question

20) Consider the following reaction at equilibrium. What effect will decreasing the pressure of the 20) reaction mixture have on the system? 2 H25g) 302(8) 2 H20(g)+2S02(6) A) The reaction will shift to the left in the direction of reactants. B) The reaction will shift to the right in the direction of products C) The equilibrium constant will decrease. D) No effect will be observed. 21) Consider the following reaction 21) COCI2(g) CO(G)+Cl2(3) A reaction mixture initially contains 1.6 MCOCI2. Determine the equilibrium concentration of CO if Ke-833×10-4. Calculate this using the approximation method. A) 2.1 x 10-2 M B) 3.7 × 10-2 M C) 4.2 × 10-4 M D) 1.3 x 10-3 M

Explanation / Answer

(20)

According to Le-chatlier's principle decrease in pressure favours the reaction towards more number of gaseous substances.

In the given equilibrium,

number of gaseous reactants = 2 + 3 = 5

Number of gaseous products = 2 + 2 = 4

SO, nP < nR

Therefore,

(B) The reaction will shift to right in the direction of products.

(21)

COCl2 = CO + Cl2 (g)

At euilibrium,

[COCl2] = (1.6 - x) M

[CO] = [Cl2] = x M

K = [CO][Cl2] / [COCl2]

8.33 * 10-4 = x * x / (1.6 - x)

0.00133 - 0.000833 x = x2

x2 + 0.000833 x - 0.00133 = 0

Using quadratic equation,

x = [ - 0.000833 +/- sqrt. ((0.000833)2 - 4(1)(-0.00133))] / 2(1)

x = 0.0360 M

Therefore, at equilibirum,

[CO] = 0.0360 M

[CO] = 3.60 * 10-2 M

So, (B)

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