Question
Define: $(a)$ Reversible process $(b)$ Standard enthalpy of combustion.
Calculate the enthalpy change for the reaction: $N _{2( g )}+3 H _{2( g )} \longrightarrow 2 NH _{3( g )}$.
The bond enthalpies are:
Bond $N \equiv N$ $H-H$ $N-H$
$\Delta H ^0$ in $kJ mol ^{-1}$ $946$ $435$ $389$

Answer

standard enthalpy of combustion:
The standard enthalpy of combustion of a substance is the standard enthalpy change accompanying a reaction in which one mole of the substance in its standard state is completely oxidised.
Reversible processes:
A thermodynamic process can be considered reversible only if it possible to retrace the path in the opposite direction in such a way that the system and surroundings pass through the same states as in the initial, direct process.
calculate the enthalpy change:
$ 3 \times(435)+946-26 \times 389$
$= 1305+946-2334$
$=-83$

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