Chemical Thermodynamics — Chemistry STD 12 Science — Question
Maharashtra BoardEnglish MediumSTD 12 ScienceChemistryChemical Thermodynamics3 Marks
Question
Obtain the expression for work done in chemical reaction.
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Answer
Consider $n_1$ moles of gaseous reactants $A$ of volume $V_1$ change to $n_2$ moles of gaseous products $B$ of volume $V_2$ at temperature $T$ and pressure P. ${ }_{V_1}^{n_1} A _{( g )} \stackrel{T}{\longrightarrow}{ }_{V_2}^{n_2} B _{( g )}$ In the initial state, $P V_1=n_1 R T$ In the final state, $P V_2=n_2 R T$ $P V_2-P V_1=n_2 R T-n_1 R T=\left(n_2-n_1\right) R T=\Delta n R T$ where $\Delta n$ is the change in number of moles of gaseous products and gaseous reactants. Due to net changes in gaseous moles, there arises change in volume against constant pressure resulting in mechanical work, $-P \Delta V$. $\therefore W =- P \Delta V =- P \left( V _2- V _1\right)=-\Delta nRT$ (i) If $n_1-n_2, \Delta n=0, W=0$. No work is performed. (ii) If $n_2>n_1, \Delta n>0$, there is a work of expansion by the system and $W$ is negative.
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