2 HNO()+ Ag ()AgNos)NO2 (9)+M20( A AgNO(s)+NO2 (9)+ H2o () Use the data from thi

Question

2 HNO()+ Ag ()AgNos)NO2 (9)+M20( A AgNO(s)+NO2 (9)+ H2o () Use the data from this tlink:ThermodynamicData to Use the data the data from this link: ThermodynamicData to calculate the following for the reaction above. (a)AG® at 298 K. -594 (b) The natural logarithm (In) of the equilibrium constant at 450.oK That is, find what goes in the box: K exp (c) The Kelvin temperature at which the reaction can be at equiibrium under standard conditions, if this is possible. If it is not possible, enter o. (d) The G In k) at 450.0 K if all reactants in solution are 0.10 M, all gaseous reactants are 0.10 atm, all products in solution are 0.30 H and all gaseous products are 0.30 at

Explanation / Answer

a)

DG = Gproducts - Hreactatns

dG = (AgNO3 + No2 + h2O)- (2HNO3 + Ag)

dG = (-33.4+ 51.3 + -237.1) - (2*-80.7 + 0)

dG = -57.8 kJ/mol

b)

dG = -RT*ln(K)

K = exp(-dG/(RT))

K = exp(57800/(8.314*450))

K = exp(15.449)

answer must be 15.449

c)

find T at which it becomes possible

dH = (-124.4+ 33.2-285.8) - (2*-174.1 + 0)

dH = -28.8

dS = (140.9 + 240.1+70.0) - (2*155.6+ 42.6)

dS = 97.2

dG < 0

dH - T*dS < 0

-28800 - T*97.2 < 0

-28800 /97.2 < T

T > 0

always possible

d)

dG = dG° + RT*ln(Q)

Q = P-NO2 = 0.3

dG = dG° + RT*ln(0.3)

dG = -57800 + 8.314*450*ln(0.3)

dG = -62304.4 J/mol

dG = -62.30kJ/mol

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