Determine the enthalpy change for the reaction: N 2 H 4 (l) + 2H 2 O 2 (l)--> N

Question

Determine the enthalpy change for the reaction: N2H4(l) + 2H2O2(l)--> N2(g) + 4H2O(l) using the following thermochemical equations. N2H4(l) + O2(g) -->N2(g) + 2H2O(l)   H=-622.2kJ H2(g) + O2(g) -->H2O2(l) H=-187.8kJ H2(g) + 1/2O2(g) -->H2O(l) H=-285.8kJ Determine the enthalpy change for the reaction: N2H4(l) + 2H2O2(l)--> N2(g) + 4H2O(l) using the following thermochemical equations. N2H4(l) + O2(g) -->N2(g) + 2H2O(l)   H=-622.2kJ H2(g) + O2(g) -->H2O2(l) H=-187.8kJ H2(g) + 1/2O2(g) -->H2O(l) H=-285.8kJ H2(g) + O2(g) -->H2O2(l) H=-187.8kJ H2(g) + 1/2O2(g) -->H2O(l) H=-285.8kJ

Explanation / Answer

N2H4(l) + O2(g) -->N2(g) + 2H2O(l)   : H 1 =-622.2kJ    ----(1 ) H2(g) + O2(g) -->H2O2(l)   : H 2 =-187.8kJ             ---- ( 2 ) H2(g) + 1/2O2(g) -->H2O(l)   : H 3 =-285.8kJ         ----- (3) N2H4(l) + 2H2O2(l)--> N2(g) + 4H2O(l)   : H= ?   ---- ( 4 ) Eq ( 4 ) can be obtained by the following operation Eq ( 4 ) = Eq ( 1 ) + 2 * reverse of Eq ( 2 ) + 2 * Eq (3 ) So , H = H1 + 2 * ( -H2 ) + 2 *H3             = -622.2kJ    + 2 * 187.8kJ   * 2 * (-285.8kJ )             = - 818.2 kJ H2(g) + O2(g) -->H2O2(l)   : H 2 =-187.8kJ             ---- ( 2 ) H2(g) + 1/2O2(g) -->H2O(l)   : H 3 =-285.8kJ         ----- (3) N2H4(l) + 2H2O2(l)--> N2(g) + 4H2O(l)   : H= ?   ---- ( 4 ) Eq ( 4 ) can be obtained by the following operation Eq ( 4 ) = Eq ( 1 ) + 2 * reverse of Eq ( 2 ) + 2 * Eq (3 ) So , H = H1 + 2 * ( -H2 ) + 2 *H3             = -622.2kJ    + 2 * 187.8kJ   * 2 * (-285.8kJ )             = - 818.2 kJ

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