You have one mole of Hf . Given that: Cp()= 23.4+ 9.18e-3 * T - 3.74e4 /T^2 - 6.

Question

You have one mole of Hf . Given that:

Cp()= 23.4+ 9.18e-3 * T - 3.74e4 /T^2 - 6.31e-7 * T^2

Cp()= 14.38 + 9.60e-3 * T + 4.95e5 /T^2 - 6.28e-8 * T^2

Cp(l)= 37.656 T(-)= 2054K

T(-l)= 2500K

S_0= 43.56 J/molK

H(-)= 5.908 kJ/mol

H( -l)= 29.288 kJ/mol

what is the enthalpy change as it is heated from T1= 2100K to T2= 3300K? P.S.: Hf has two solid phases (, and ). The two temperatures are the temperatures of the phase transitions. S_0 is the standard entropy at 298K. The enthalpies are the latent heats of the transitions at the temperature of transition.

a. 73.9 kJ

b. 34.5 kJ

c. 65.3 kJ

d. 92826 J

e. 65.23 kJ

f. 4.28 kJ

g. 47.7 kJ

Explanation / Answer

Given Cp(alpha) = 23.4 + 9.81*10^-3*T - 3.7*10^4/T^2 - 6.31*10^-7*T^2
Given Cp(beta) = 14.38 + 9.6*10^-3*T + 4.95*10^5/T^2 - 6.28*10^-8*T^2
Cp(l) = 37.656

T(alpha-beta) = 2054 K
T(beta-l) = 2500 k

S_o = 43.56 J/mol K
dH(alpha-beta) = 5.908*10^3 J/mol
dH(beta-l) = 29.288*10^3 J/mol

so, enthalapy change from T1 = 2100 K to T2 = 3300 K
dH = [integrate]Cp(beta)dT (from 2100K to 2500K) + [integrate]Co(l)dT[from 2500 to 3300 K] + dH(beta-l)
dH = [integrate](14.38 + 9.6*10^-3*T + 4.95*10^5/T^2 - 6.28*10^-8*T^2)dT + 37.656(3300 - 2500) + 29.288*10^3
dH = (14.38(2500 - 2100) + 9.6*10^-3*(2500^2 - 2100^2)/2 - 4.95*10^5(1/2500 - 1/2100) - 6.28*10^-8*(2500^3 - 2100^3)/3) + 37.656(3300 - 2500) + 29.288*10^3
dH = 73901.294 J = 73.901 kJ

option a.

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