The enthalpies of three reactions at 25 Celsius are given inKJ 2H2 (g) + O2(g)..

Question

The enthalpies of three reactions at 25 Celsius are given inKJ      2H2 (g) + O2(g).........2H2O(l)      -572 H in KJ/mol      2C2H6 (g) + 702 (g)......4CO2(g) +6H2O (l)    -3120 H in kJ/mol      C2H4(g) 3O2(g).........2CO2(g) + 2H2O(l)       -1411 H inkJ/mol Use this information to calculate the change in enthalpyfor C2H6(g)..........C2H4(g) + H2(g) Is the reaction endothermic or exothermic? explain The enthalpies of three reactions at 25 Celsius are given inKJ      2H2 (g) + O2(g).........2H2O(l)      -572 H in KJ/mol      2C2H6 (g) + 702 (g)......4CO2(g) +6H2O (l)    -3120 H in kJ/mol      C2H4(g) 3O2(g).........2CO2(g) + 2H2O(l)       -1411 H inkJ/mol Use this information to calculate the change in enthalpyfor C2H6(g)..........C2H4(g) + H2(g) Is the reaction endothermic or exothermic? explain

Explanation / Answer

2C2H6 (g) + 702 (g)......4CO2(g) + 6H2O(l)    -3120 H inkJ/mol      eq2 C2H4(g) 3O2(g).........2CO2(g) + 2H2O(l)       -1411 H inkJ/mol      eq3 adding 1/2 ( eq2) - eq3 - 1/2 ( eq 1)       we get the desiredreaction as: C2H6(g)--->C2H4(g) + H2(g) so the change in enthalpy is 1/2 ( H2) - H3 - 1/2 (H1) => 1/2 ( -3120 KJ /mol) + 1411 KJ/mol + 1/2 ( 572KJ/mol) => 156.72 KJ/mol as here net change in enthalpy is positive so reaction isendothermic i.e heat is absorbed.   

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