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Bond energy of `(N-H)` bond is `yKJ mol^(-1)` under standard state. Thus,change in internal energy in the following process is `NH_(3)(g)rarrN(g)+3H(g)`A. `-3yKJmol^(-1)`B. `-yKJmol^(-1)`C. `3yKJmol^(-1)`D. `yKJmol^(-1)` |
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Answer» Correct Answer - D `H-underset(H)underset(|)(N)-H` (three `N-H`) bonds Thus, `DeltaE=3yKJmol^(-1)` |
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