At 298 K the standard enthalpy of combustion of solid L-tyrosine ((C 9 H 11 NO 3

Question

At 298 K the standard enthalpy of combustion of solid L-tyrosine ((C9H11NO3 (s)) to give CO2 (g), H2O (liq) and N2 (g) is combH° = –4428 kJ/mol. Calculate rxH° and rxU° at 298 K for the following reaction.

C9H11NO3 (s) + 17 H2 (g) 9 CH4 (g) + NH3 (g) + 3 H2O (liq)

rxH° = –4155 kJ; rxU° = –4138 kJ

rxH° = 1323 kJ; rxU° = 1340 kJ

rxH° = –891.3 kJ; rxU° = 1.645x104 kJ

rxH° = 279.4 kJ; rxU° = 7623 kJ

rxH° = 279.4 kJ; rxU° = 296.7 kJ

rxH° = 2851 kJ; rxU° = 2.020x104 kJ

rxH° = –4155 kJ; rxU° = 1.319x104 kJ

rxH° = –891.3 kJ; rxU° = –874.0 kJ

rxH° = 1323 kJ; rxU° = 1.867x104 kJ

rxH° = 2851 kJ; rxU° = 2869 kJ

a.

rxH° = –4155 kJ; rxU° = –4138 kJ

b.

rxH° = 1323 kJ; rxU° = 1340 kJ

c.

rxH° = –891.3 kJ; rxU° = 1.645x104 kJ

d.

rxH° = 279.4 kJ; rxU° = 7623 kJ

e.

rxH° = 279.4 kJ; rxU° = 296.7 kJ

f.

rxH° = 2851 kJ; rxU° = 2.020x104 kJ

g.

rxH° = –4155 kJ; rxU° = 1.319x104 kJ

h.

rxH° = –891.3 kJ; rxU° = –874.0 kJ

i.

rxH° = 1323 kJ; rxU° = 1.867x104 kJ

j.

rxH° = 2851 kJ; rxU° = 2869 kJ

Explanation / Answer

Answer:

Using below mentioned it is possible to get required,

Hr = Ur +pV

Hr = HC9H11NO3 + HH2-HCH4 - HNH3 - HH2O

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