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The enthalpies of formation of \( N _{2} O \) and NO at \( 298 K \) are 82 and \( 90 kJ mol { }^{-1} \). The enthalpy of the reaction : \( N _{2} O ( g )+1 / 2 O _{2}( g ) \longrightarrow 2 NO ( g ) \) is (A) \( -8 kJ \) (B) \( 98 kJ \) (C) \( -74 kJ \) (D) \( 8 kJ \). |
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Answer» \(\Delta H_r = \Delta H_{NO}-(\Delta H_{N_2O} + \Delta H_{O_2} )\) = 90KJ x 2 - (82 + 0) = 180 - 82 = 98 KJ |
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