2KClO4(s) 2 KClO3(s) + O2(g) Delta S degree rxn = J/K Delta S degree surr = J/K

Question

2KClO4(s) 2 KClO3(s) + O2(g) Delta S degree rxn = J/K Delta S degree surr = J/K Is the reaction spontaneous at 25 degree C? yes no 2Na(s) + 2H2O(l) rightarrow 2NaOH(aq) + H2(g) Delta S degree rxn = J/K Delta S degree surr = times 10 J/K (enter your answer in scientific notation) Is the reaction spontaneous at 25degreeC? yes no N2(g) rightarrow 2N(g) Delta S degree xn = J/K Delta S degree surr = times10 J/K(enter your answer in scientific notation) Is the reaction spontaneous at 25degree C? yes no

Explanation / Answer

a)

dH = Hproducts - Hreactants

dH = (2*KClO3 + O2) - (2*KClO4)

dH = (2*-397.7+ 0) - (2*-430.1152)

dH = 64.8304 kJ/mol

then,

dSsurroundings = Qsurr/T = -Hsystem / T = -64.8304 kJ/mol / 298 K = 0.21755kJ/molk = -217.55 J/molK

dSrxn = dSproducts - dSreactants

dSrxn =  (2*KClO3 + O2) - (2*KClO4)

dSrxn =  (2*142.96728+ 205.2) - (2*151.0424)

dSrxn = 189.04 J/molK

the reaction spontaneously:

dSuniverse = dSsystem + dSsurrounding

dSuniverse = 189.04+ 217.55 = -28.51 J/molK

this decreases, therefore, it must NOT be spontaneous

b)

dSrxn = 2NaOH + H2 - (2NA + 2H2O)

dSrxn = 2*48.25 + 130.7 - (2*51.3 + 2*70.0) = -15.4

for surroundings:

dSsurr = Qsurr/Tsurr = -Hsys/T

Hsys = (2*-470.09 + 0) - (2*0 + 2*-285.8) = -368.58 kJ/mol

dSsurr = Qsurr/Tsurr = -Hsys/T = -(-368.58 *10^3)/(298) = 1236.84 J/molK

This is likely to occur, since Entropy increases

C)

N2(g) = 2N(g)

dSsystem = 2*153.188792 - 1*191.50168 = 114.87J/molK

Hsyst = 2*472.704136 - 0 = 945.408 kJ/mol

dSsurr = -Hsyst/T = -945.408*1000/298 = -3172.510 J/molK

Now,

dSuiversE = -3172.510 + 114.87

dsuniverse = -3057.64 J/molK

at STP, not likely to occur

c)

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