Question

Correct mathematical formulation Of First Law Of Thermodynamics used In thermochemistry is

• A. $∆\mathrm{H}=∆\mathrm{U}-∆\mathrm{nRT}$
• B. $∆\mathrm{U}=∆\mathrm{H}+∆\mathrm{nRT}$
• C. $∆\mathrm{H}=∆\mathrm{U}+∆\mathrm{nRT}$
• D. $∆\mathrm{H}=-∆\mathrm{U}+∆\mathrm{nRT}$

Answer 1

The correct option is C

$∆\mathrm{H}=∆\mathrm{U}+∆\mathrm{nRT}$

Explanation for correct options

(c) $∆\mathrm{H}=∆\mathrm{U}+∆\mathrm{nRT}$

• Let ${\mathrm{H}}_1,{\mathrm{U}}_1,{\mathrm{V}}_1\mathrm{and}{\mathrm{H}}_2,{\mathrm{U}}_2,{\mathrm{V}}_2$​ represent enthalpies, internal energies, pressures and volumes in the initial and final states respectively.
• For a reaction involving ${\mathrm{n}}_1$​ moles of gaseous reactants in initial state and ${\mathrm{n}}_2$​ moles of gaseous products at final state,

${\mathrm{n}}_1{\mathrm{X}}_{\left(\mathrm{g}\right)}\to {\mathrm{n}}_2{\mathrm{Y}}_{\left(\mathrm{g}\right)}$

• If ${\mathrm{H}}_1\mathrm{and}{\mathrm{H}}_2$​ are the enthalpies in initial and final states respectively, then the heat of reaction is given by enthalpy change as

$∆\mathrm{H}={\mathrm{H}}_2-{\mathrm{H}}_1$

• Mathematical definition of $\mathrm{H}\mathrm{is}\mathrm{H}=\mathrm{U}+\mathrm{PV}$
• $$\begin{gathered} \mathrm{Thus},{\mathrm{H}}_1={\mathrm{U}}_1+{\mathrm{P}}_1{\mathrm{V}}_1\mathrm{and}{\mathrm{H}}_2={\mathrm{U}}_2+{\mathrm{P}}_2{\mathrm{V}}_2 \\ \Rightarrow ∆\mathrm{H}={\mathrm{U}}_2+{\mathrm{P}}_2{\mathrm{V}}_2-({\mathrm{U}}_1+{\mathrm{P}}_1{\mathrm{V}}_1) \\ \Rightarrow ∆\mathrm{H}={\mathrm{U}}_2+{\mathrm{P}}_2{\mathrm{V}}_2-{\mathrm{U}}_1-{\mathrm{P}}_1{\mathrm{V}}_1 \\ \Rightarrow ∆\mathrm{H}={\mathrm{U}}_2-{\mathrm{U}}_1+{\mathrm{P}}_2{\mathrm{V}}_2-{\mathrm{P}}_1{\mathrm{V}}_1 \\ \mathrm{where},∆\mathrm{U}={\mathrm{U}}_2-{\mathrm{U}}_1 \end{gathered}$$
• $$\begin{gathered} \mathrm{Now},\mathrm{since}\mathrm{PV}=\mathrm{nRT} \\ \mathrm{For}\mathrm{initial}\mathrm{state},{\mathrm{P}}_1{\mathrm{V}}_1={\mathrm{n}}_1\mathrm{RT} \\ \mathrm{For}\mathrm{final}\mathrm{state},{\mathrm{P}}_2{\mathrm{V}}_2={\mathrm{n}}_2\mathrm{RT} \\ \mathrm{Subtracting}\mathrm{initial}\mathrm{and}\mathrm{final}\mathrm{state} \\ {\mathrm{P}}_2{\mathrm{V}}_2-{\mathrm{P}}_1{\mathrm{V}}_1={\mathrm{n}}_2\mathrm{RT}-{\mathrm{n}}_1\mathrm{RT} \\ =({\mathrm{n}}_2-{\mathrm{n}}_1)\mathrm{RT} \\ =∆\mathrm{nRT} \end{gathered}$$
• Hence, $∆\mathrm{H}=∆\mathrm{U}+∆\mathrm{nRT}$
• Therefore the correct option is (c) $∆\mathrm{H}=∆\mathrm{U}+∆\mathrm{nRT}$.

Explanation for incorrect options

Since, the correct formula is $∆\mathrm{H}=∆\mathrm{U}+∆\mathrm{nRT}$. Therefore, options (a), (b) and (d) are incorrect.

Therefore the correct option is (c) $∆\mathrm{H}=∆\mathrm{U}+∆\mathrm{nRT}$