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Given the bond energirs N=N,H-H and N-H ...

Given the bond energirs `N=N,H-H` and `N-H` bonds are `945,436` and `391KJmol^(-1)` respectively, the enthalpy of the follwing reactlon `N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g)` is

A

`-93`kJ

B

102 kJ

C

90 kJ

D

105 k

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The correct Answer is:
A
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Given the bond energies N=N , H and H-H bond are 945, 436 and 391KJmol^(-1) respectively, the enthalpy change of the reaction N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g) is

For the reaction, N_(2)(g)+3H_(2)(g) to 2NH_(3)(g)

For the reaction N_(2)(g) + 3H_(2)(g) hArr 2NH_(3)(g), DeltaH=?

The standard enthyalpy of N_(2(g))+3H_(2(g)rarr2NH_(3(g)) is

Which of the following is correct for the reaction? N_(2)(g) + 3H_(2)(g) rarr 2NH_(3)(g)

TARGET PUBLICATION-CHEMICAL THERMODYNAMICS AND ENERGETICS -CRITICAL THINKING
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