Calculate Delta S degree for the evaporation of water, given S degree [H_2O(l)]

Question

Calculate Delta S degree for the evaporation of water, given S degree [H_2O(l)] = + 69.9 J/mol^-1 K and S degree [H_2O(g)] = + 188.7 J/mol^-1 K. - 258.8 J/K - 118.8 J/K 0.0000 J/K + 118.8 J/K + 258.6 J/K For which of the following reactions will the entropy of the system decrease? 2NH_3(g) rightarrow N_2(g) + 3H_2(g) 2C(s) + O_2(g) rightarrow 2CO(g) CaCO_3(s) rightarrow CaO(s) + CO_2(g) 2NO_2(g) rightarrow N_2O_4(g) NaOH(s) rightarrow Na*(aq) + OH^- (aq) All of the following statements are true EXCEPT A reaction is spontaneous if Delta G  0, then a reaction is not spontaneous. Delta G is referred to as Gibbs free energy.

Explanation / Answer

28 )

evaporation of water

H2O (l) -----> H2O (g)

delta S = S of H2O(g) - S of H2O (l)

delta S = 188.7 - 69.9 = + 118.8 J/K ( option d)

29)

entropy of gas >>> liquid >> solid

if product contains more gaseous species than reactant , entropy of system increases and vice versa

option d) 2 NO2 (g) ---> N2O4 (g) more gaseous species in reactant side so entropy of the system decreases.

30) dG = dH - TdS

if dG is negative reaction spontaneous

if dG is positive reaction is not spontaneous

if dG = 0 reaction will be in equlibrium

at dG = 0 , dH = -TdS

so option c) if dG = 0 , dS = dH is incorrect.

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