Calculate the change in enthalpy for thefollowing reaction, using the bond enthalpydata provided below.(note: you may roundthe bond energies to the nearest 100kj·mol−1during your calculations)2n2h2(g) + 3o2(g)←→4no(g) + 2h2o(g)bondbond energy (kj·mol−1)

There are missing information in the question. The bonds enthalpy are:

N=N: 409 kJ/mol

N-H: 388 kJ/mol

O=O: 498 kJ/mol

N=O: 630 kJ/mol

O-H: 463 kJ/mol

+508 kJ/mol

Explanation:

The formation of a bond is endothermic, so the enthalpy is positive, and the break is exothermic, so the enthalpy is negative. The enthalpy of the reactants must be negative, thus, the change in enthalpy can be calculated as:

ΔH = ∑n*H products - ∑n*H reactants, where n is the number of bonds.

N₂H₂ has 1 N=N bond and 2 N-H bonds; O₂ has 1 O=O bond; NO has 1 N=O bond; and H₂O has 2 O-H bonds, so:

ΔH = (4*1*630 + 2*2*463) - [2*(1*409 + 2*388) + 3*1*498)

ΔH = +508 kJ/mol

oxygen gas properties are as follow:

Explanation:

oxygen gas is colourless,odorless,tasteless gas.It changes from a gas to liquid at the temperature of -182.96°C.The liquid form has a slightly blush colour to it.Liquid oxygen can also be frozen at the temperature of -218.4°C.