Solveeit Logo
Login

Question

Question: For the reaction: \[2{{A}_{(g)}}+3{{B}_{(g)}}\to 4{{C}_{(g)}}+{{D}_{(l)}}\] ; \[\Delta H=-20kJ/mole\...

For the reaction: 2A(g)+3B(g)→4C(g)+D(l)2{{A}_{(g)}}+3{{B}_{(g)}}\to 4{{C}_{(g)}}+{{D}_{(l)}} ; ΔH=−20kJ/mole\Delta H=-20kJ/mole. Find the heat evolved when 0.40.4 mole of A reacts with an excess of B in a closed rigid container. The temperature is constant at 300K300K. [R=8.3J/Kmole][R=8.3J/Kmole]

Explanation

Solution

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 22 moles of A and 33 moles of B on reaction gives, 44 moles of C and 11 of D. But in question we have only 0.40.4 moles of A. According to this, we have;
No. of moles of C when 22 moles of A reacts =44moles
No. of moles of C when 0.40.4 moles of A reacts = 0.80.8 moles
No. of moles of B =33 (As B is in excess amount ,given)
Hence, we have total moles of reactant(moles of A and moles of B)= 0.4+3=3.40.4+3=3.4
Total moles of product = 0.80.8
Hence, we have known the equation of enthalpy of reaction that is given below;
ΔH=qv+ΔngRT\Delta H={{q}_{v}}+\Delta {{n}_{g}}RT
ΔH=\Delta H= Enthalpy of reaction, ΔH=−20kJ/mole\Delta H=-20kJ/mole
qv={{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=R=Ideal gas constant, R=8.3J/KmoleR=8.3J/Kmole (Given i)
T=T=Temperature of reaction, T=300KT=300K
Let us substitute these value in the above equation for the enthalpy of reaction;
ΔH=qv+ΔngRT\Delta H={{q}_{v}}+\Delta {{n}_{g}}RT
−20×103=qv+(−3×8.3×300)-20\times {{10}^{3}}={{q}_{v}}+(-3\times 8.3\times 300)
⇒qv=−12530J/mol=−12.53KJ/mol\Rightarrow {{q}_{v}}=-12530J/mol=-12.53KJ/mol

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

Note: Here the change in enthalpy is given in the unit of kJmol−1kJmo{{l}^{-1}}.So, we have to convert this value into Jmol−1Jmo{{l}^{-1}} by multiplying the value with 103{{10}^{3}} for the calculation. According to the second law of thermodynamics entropy of the universe always increases for a spontaneous process.