Q.

Use the information in the table to calculate the enthalpy of this reaction.

C2H6(g)+72O2(g)2CO2(g)+3H2O(l)

 

ReactionHf° kJ/mole
2C(s) + 3 H2(g)C2H6(g)-84.7
2C(s) + O2(g) CO2(g)-393.5
H2(g)12O2(g) H2O(l)-285.8

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a

-1664 kJ

b

-1560 kJ

c

- 764 kJ

d

-3120 kJ

answer is B.

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Detailed Solution

The change in the enthalpy of this reaction is,

ΔH=393.5×2285.8×3+84.7 =1560kJ

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