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When NH4NO2(s), decomposes at 373 K, it forms N2 (g) and H2O (g). The ∆H for the reaction at one atmospheric pressure and 373 K is −223. 6 kJ mol -1 . What is the value of ∆E for the reaction under the same conditions? (Given R = 8.31 JK-1 mol-1) |
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Answer» NH4NO2(S) → N2 (g) + 2H2O (g), ∆ng= 3 − 0 = 3 ∆U = ∆H − ∆ng RT = −223.6 kJ mol−1 − (3 mol) x (8.314 × 10−3 kJ K−1 mol−1) × (373 K) = −232.9 kJ mol−1 . |
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