The molar solubility of Fe(OH), in a 0.200 M FeCI solution is. K_ap = 4.00 times

Question

The molar solubility of Fe(OH), in a 0.200 M FeCI solution is. K_ap = 4.00 times 10^-10 a) 7.41 times 10^-19 b) 5.86 times 10^-13 c) 1.95 times 10^-13 d) 2.02 times 10^-13 Indicate some other answer For a reaction where Delta H is positive and Delta S is positive, the reaction will a) spontaneous at low temperature b) spontaneous at all temperatures. c) spontaneous at high temperatures. d) nonspontaneous at high temperatures.. nonspontaneous at all temperatures. Based on the relationship of entropy to the degree of disorder in a the following changes are accompanied by a decrease in entropy? I. The extraction of metal from their ores. II. A glassblower heating glass to its softening temperature. III. The sublimation of iodine. IV. The formation of stalactites and stalagmites in caves. V. 2 N_2 (g) + 4H_2 O(g) + O_2 (g) rightarrow 2 NH_4 NO_3 (s) a) III, IV, and V b) I, IV and V c) I and IV E) II, III, and V for the reaction, IF_s (g) rightarrow IF_3 (g) + F_2 (g), is kJ, given the IF(g) + F_2 (g) rightarrow IF_3 (g) -390 IF(g) + 2 F_2 (g) rightarrow IF_3 (g) -745 a) -1135 b) + 1135 c) -35 d) + 35

Explanation / Answer

Fe(OH)3---------->Fe+ 3 + 3OH-

KSp = [Fe+3] [OH-]3

Given 0.2M FeCl3 suppliments Fe+3 of 0.2M

Let x= molar solubility of Fe(OH)2

4*10-38= (x+0.2)* x3

When solved using excel x= 5.86*10-13M ( b is correct)

2. For a reaction to be spontaneous, deltaG= deltaH- Tdelta S has to be negative,

since deltaS and deltaH are +ve. deltaG is -ve if T deltaS> deltaH

T > deltaH/deltaS this is possible at very high temperatures. ( C is correct)

3. Entropy is degrees of disorder. entropy of solids is less than liquids and in turn less than gases. Hence for reaction (choice V. entropy decreases since gases becamee solids, Sublimation refer to solid becoming gas, so there will be an increase in entropy. Softening decreases entropy, the formation of stalacites also decreases entropy. Hence i,IV and V are correct choices.

4.

IF +F2--------->IF3   deltaG= -390 KJ (1)

Reversing the second reaction give IF5------>iF + 2F2   deltaG= 745 KJ (2)

Addition of 1 and 2 gives IF5 --->IF3+F2 deltaG= 745-390= 355 KJ

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