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the reaction of nitrogen monoxide(g) with oxygen(g) to form nitrogen dioxide(g) proceeds as follows: 2 no(g) o2(g) 2 no2(g) when 7.62 grams of no(g) react with sufficient o2(g) , 14.5 kj of energy are evolved . what is the value of h for the chemical equation given?



Answer :

The value of enthalpy change for the chemical equation given when 7.62 grams of no(g) react with sufficient o2(g) , 14.5 kj of energy are evolved is 114.16 kJ

The reaction is 2NO(g)+O2(g)---2NO2(g)

7.62 grams of NO(g) reacts with sufficient O2(g), 14.5 kJ of energy is evolved.

So NO(g) is the limiting reagent.

Moles of NO given = 7.62/30=0.254

So 0.254 mole of NO reacts to release 14.5 kJ energy

2 mole of NO will release 14.5/0.254x2=114.16 kJ

In a balanced reaction, 2 mole of NO reacts

So deltaH(rxn)= 114.16 kJ

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