Question

For the reaction: \[2{{A}_{(g)}}+3{{B}_{(g)}}\to 4{{C}_{(g)}}+{{D}_{(l)}}\] ; \[\Delta H=-20kJ/mole\]. Find the heat evolved when \[0.4\] mole of A reacts with an excess of B in a closed rigid container. The temperature is constant at \[300K\]. \[[R=8.3J/Kmole]\]

Answer 1

Hint: The Heat of Reaction (also known and Enthalpy of Reaction) is the change in the enthalpy of a chemical reaction that occurs at a constant pressure. It is a thermodynamic unit of measurement useful for calculating the amount of energy per mole either released or produced in a reaction.

Complete step by step answer:
Here, we can see that when \[2\] moles of A and \[3\] moles of B on reaction gives, \[4\] moles of C and \[1\] of D. But in question we have only \[0.4\] moles of A. According to this, we have;
No. of moles of C when \[2\] moles of A reacts =\[4\]moles
No. of moles of C when \[0.4\] moles of A reacts = \[0.8\] moles
No. of moles of B =\[3\] (As B is in excess amount ,given)
Hence, we have total moles of reactant(moles of A and moles of B)= \[0.4+3=3.4\]
Total moles of product = \[0.8\]
Hence, we have known the equation of enthalpy of reaction that is given below;
\[ \Delta H={{q}_{v}}+\Delta {{n}_{g}}RT\]
\[\Delta H=\] Enthalpy of reaction, \[\Delta H=-20kJ/mole\]
\[{{q}_{v}}= \] Heat involved in the reaction (that is we have to find)
\[\Delta {{n}_{g}}\Rightarrow \]\[{{n}_{P}}-{{n}_{R}}=0.8-3.8=-3\]
\[R=\]Ideal gas constant, \[R=8.3J/Kmole\] (Given i)
\[T=\]Temperature of reaction, \[T=300K\]
Let us substitute these value in the above equation for the enthalpy of reaction;
\[ \Delta H={{q}_{v}}+\Delta {{n}_{g}}RT\]
\[ -20\times {{10}^{3}}={{q}_{v}}+(-3\times 8.3\times 300)\]
\[\Rightarrow {{q}_{v}}=-12530J/mol=-12.53KJ/mol\]

Here we got the heat involved in reaction is equal to \[-12530J/mol\] or \[-12.53KJ/mol\].

Note: Here the change in enthalpy is given in the unit of \[kJmo{{l}^{-1}}\].So, we have to convert this value into \[Jmo{{l}^{-1}}\] by multiplying the value with \[{{10}^{3}}\] for the calculation. According to the second law of thermodynamics entropy of the universe always increases for a spontaneous process.