Question

the following with a "t" ora" -?) for the following example H, Ocs-> H2000 Identity 9 Sys.at A Hsys Assys sesurr Assurr.

Answer 1

1.Ans :-

Given physical reaction is :

H₂O (s) --------------> H₂O (l)

(a). Because, the given reaction is endothermic reaction, therefore sign of heat of the system (q_sys.) would be positive (+ve) as system absorb the heat in the reaction.

(b). Because, the given reaction is endothermic reaction, therefore sign of change in enthalpy of the system (ΔH_sys.) would be positive (+ve) as system absorb the heat in the reaction.

(c). Because, in this reaction entropy increases as solid changes into liquids which have more degree of randomness, therefore sign of change in entropy of the system (ΔS_sys.) would be positive (+ve).

(d). Because, the given reaction is endothermic reaction, therefore sign of change in enthalpy of the surrounding (ΔH_surr..) would be negative (-ve) as surrounding evolve the heat in the reaction.

(e). Because, ΔS_surr.= -ΔH/T, therefore sign of change in entropy of the surrounding (ΔS_surr.) would be negative (-ve).

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2.Ans :-

Let solubility of Mn(OH)₂ in pure water = S mol/L

Partial dissociation of Mn(OH)₂ is :

Mn(OH)₂ (s) <------------------> Mn²⁺ (aq) + 2OH^- (aq)

...................................................S mol/L........2S mol/L

Expression of solubility product i.e. K_sp(which is equal to the product of the molar concentration of products raise to power their stoichiometric coefficient when reaction is at equilibrium stage).

K_sp = [Mn²⁺].[OH^-]²

4.30 x 10^-14 = S(2S)²

4S³ = 4.30 x 10^-14

S = (4.30 x 10^-14 /4)^1/3

S = (10.75 x 10^-15)^1/3

S = 2.21 x 10^-5 mol/L

Therefore,

[Mn2+] = S = 2.21 x 10-5 M

and

[OH-] = 2S = 2 x 2.21 x 10-5 M = 4.42 x 10-5 M