The equilibrium constant for the acid ionization of (HSCH_2CH_2OH) is 1.91 times

Question

The equilibrium constant for the acid ionization of (HSCH_2CH_2OH) is 1.91 times 10^2 HSCH_2CH_2OH (aq) H^2(aq) + SCH_2CH_2OH^- (aq) Which of the following is true regarding this equilibrium? the reaction is product favored. The reaction is reactant favored. Equilibrium lies far to the right. Equilibrium lies far to the left. I and III I and IV II and III II and IV None are true, as the concentrations of reactants and products are comparable. Increasing the temperature of an exothermic reaction results in more products and fewer reactants. more reactants and fewer products. more reactants and products. fewer reactants and products. no change in the quantities of reactants and products.

Explanation / Answer

Q.34: deltaS will be negative when the entropy of all the products formed is less than the entropy of all the reactants i.e. entropy decreases from reactant to product.

Among the given options, in option (a), H2O(g) having higher entropy is converted to H2O(s) having lower entropy. Hence entropy decreases from reactant to product.

Hence deltaS is negative for option (a).

Q.35: Moles of Al2S3 taken for reaction = mass of Al2S3 / molar mass of Al2S3 = 65.0 g / 150.2 g/mol

= 0.4328 mol Al2S3(s)

From the balanced reaction it is clear that 1 mol of Al2S3(s) produces 3 mol of H2S(g).

=> 0.4328 mol Al2S3(s) that will produce the moles of H2S(g) = 0.4328 mol Al2S3 x (3 mol H2S / 1mol Al2S3)

= 1.2983 mol H2S(g)

Hence moles of H2S(g) produced, n = 1.2983 mol

Pressure, P = 0.987 atm

Temperature, T = 90.0 DegC = 90.0 + 273.15 = 363.15 K

Volume, V = ?

Applying ideal equation

PV = nRT

=> V = nRT / P

=> V = (1.2983 mol x 0.0821L.atm.mol-1K-1 x 363.15 K) / 0.987 atm

=> V = 39.2 L (answer)

Hence option (a) is correct.

Q.36: Given the equilibrium constant, K = 1.91x10-10  

Since the value of K is much less than 1, the reaction is reactant favored and the equilibrium liesfar to the left.

Hence both II and IV are correct.

Hence option (d) is correct.

Q.37: Since heat is released during an exothermic reaction, decreasing temperature favors the reaction towards product and increasing temperature favors the reaction towards reactant.

Hence more reactant and fewer product are formed.

Hence option (b) is correct

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