N_(2(g)) + O_(2(g)) + 180.6 kJ to 2NO_((g)), calculate (a) heat of reaction, (b) heat of formation of nitric oxide and (c) heat required to form one litre of nitric oxide at 25^@C.

N_(2(g)) + O_(2(g)) + 180.6 kJ to 2NO_((g)), calculate (a) heat of rea

1.	N_(2(g)) + O_(2(g)) + 180.6 kJ to 2NO_((g)), calculate (a) heat of reaction, (b) heat of formation of nitric oxide and (c) heat required to form one litre of nitric oxide at 25^@C.
Answer» Solution :THERMOCHEMICAL equation, `N _(2 (g)) + O _(2(g)) to 2 NO _((g )), Delta H =+ 180.6 kJ.`The heat of the given reaction `=Delta H = + 180.6 kJ`Since 2 mole of nitric acid is formed in the reaction, the heat of formation of nitric oxide,`Delta H _(F ) = (Delta H)/(2) = (180.6)/(2) = 90.3 kJ mol ^(-1)` At `25^(@)C` and 1 ATM ONE mole of a gas occupies 24.4 LITRES. Heat required to form 24.4 litre of NO = 90.3 kJ . Heat required to form one litre of NO `= (90.3)/(24.4) = 3.7 kJ lit ^(-1)`