Calculate the enthalpy of this reaction, based on the following Delta H_f degree

Question

Calculate the enthalpy of this reaction, based on the following Delta H_f degree values for each substance in the reaction: N_2H_4(l) + 2N_2O_4(g) rightarrow 6NO(g) + 2H_2O(l) a. -102 kJ b. 102 kJ c. -257 kJ d. 257 kJ For Question 31 show your work on this page to cam credit. Calculate the standard enthalpy of formation (Delta H_f^0) of pentane (C_5H_12(g)) as in this equation. 5C(s) + 6 H_2(g) rightarrow C_5H_12(g) Use the following standard enthalpies of combustion data to find Delta H_f^0 of pentane: C_5H_12(g) + 8O_2(g) rightarrow 5CO_2(g) + 6H_2O(g) -3271.4 J 2H_2(g) + O_2(g) rightarrow 2H_2O(g) -483.6 J C(S) + O_2(g) rightarrow CO_2(g) -393.5 J

Explanation / Answer

30)

consider the given reaction

N2H4 + 2 N204 ---> 6 NO + 2 H20

we know that

dHo rxn = dHof products - dHof reactants

so

dHorxn = ( 6 x dHof NO) + ( 2 x dHof H20) - ( dHof N2H4) - ( 2 x dHof N204)

using the values in the table

dHorxn = ( 6 x 90.4) + ( 2 x -285.8) - ( 50.6) - ( 2 x 11.1)

dHo rxn = -102 kJ

so

the enthalpy of the reaction is -102 kJ

the answer is a) -102 kJ

31)

consider the given reactions

a) C5H12 + 8 02 ---> 5 C02 + 6 H20 ----> dHa = -3271.4

b) 2 H2 + 02 ---> 2H20 ------> dHb = -483.6

c) C + 02 --> C02 ------> dHc = -393.5

d) 5C + 6H2 ---> C5H12 ------> dHd = ??

we can see that

eq d = ( 3 x eq b) + ( 5 x eq c) - ( eq a)

so

using Hess law

dHd = ( 3 x dHb) + ( 5 x dHc) - (dHa)

using given values

we get

dHd = ( 3 x -483.6) + ( 5 x -393.5) - ( -3271.4)

dHd = -146.9

so

the standard enthapy of formation of pentane gas is -146.9 J

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