Calculate DS 0 , DH 0 and DG 0 using values in Appendix G for the following reac

Question

Calculate DS0, DH0 and DG0 using values in Appendix G for the following reactions.

(a) MnO2 (s) ---> Mn(s) + O2(g)

(b) CS2(g) + 3Cl2(g) ---> CCl4(g) + S2Cl2(g)

Part B:

Identify whether the equations are spontaneous under standard state and 250C. If non-spontaneous, would high or low temperature make it spontaneous?

I know the answers are Part A: (a) DS0= +184.2 J/K, DH0= +520.03 kJ, DG0= +465.1 kJ; (b) DS0= -278.2 J/K, DH0= -232.1 kJ, DG0= -154.3 kJ; Part B: nonspontaneous but at higher temp will be spontaneous, spontaneous.

Explanation / Answer

First, recall that if

dG < 0, then this is spontaneous

if dG > this favours reactants, i.e. reverse

Also

dG = dH - T*dS

T = 25°C = 298K

for

a)

MnO2 (s) ---> Mn(s) + O2(g)

dG = dH - T*dS

dG = 520.03 -298*(184.2/1000) = 465.1384 kJ/mol

this is not favoured at 298K,

let us find T for spontaneous

dG < 0 so

dG = dH - T*dS < 0

dH - T*dS < 0

dH < T*dS

T > dH/dS = (520.03)/(184.2/1000) > 2823.18 K

T must be higher than 2823.18 in order for this equatino to be spontaneous

(b) CS2(g) + 3Cl2(g) ---> CCl4(g) + S2Cl2(g)

dG = dH - T*dS

dG = -232.1 - 298*(-278.2/1000) = -149.19 kJ/mol

so this is favoured already

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