Chemical Thermodynamics Chapter 4 Chemistry Class 12 Textbook Solution

answer:- **Given:** Enthalpy change ($\mathrm{\Delta }H$) = -620 J Volumes of reactants: $C_2{H}_4$ = 100 mL, $H_2$ = 100 mL Pressure ($P_{\text{ext}}$) = 1 bar **To find:** Pressure-volume work ($W$) and change in internal energy ($\mathrm{\Delta }U$) for the given reaction **Formulas:** 1) $W=-P_{\text{ext}}\mathrm{\Delta }V$ 2) $\mathrm{\Delta }H=\mathrm{\Delta }U+P_{\text{ext}}\mathrm{\Delta }V$ **Calculation:** According to the equation of the reaction, 1 mole of $C_2{H}_4$ reacts with 1 mole of $H_2$ to produce 1 mole of $C_2{H}_6$. Hence, 100 mL of $C_2{H}_4$ would react with 100 mL of $H_2$ to produce 100 mL of $C_2{H}_6$. $V_1$ = 100 mL + 100 mL = 200 mL = 0.2 dm${}^3$ $V_2$ = 100 mL = 0.1 dm${}^3$ From formula (1), $W=-P_{\text{ext}}\mathrm{\Delta }V$ $=-1\text{ bar}\cdot (0.1{\text{ dm}}^3-0.2{\text{ dm}}^3)$ $=0.10{\text{ dm}}^3\text{ bar}$ $=0.10{\text{ dm}}^3\text{ bar}\cdot 100\frac{\text{J}}{{\text{dm}}^3\text{bar}}=+10.00\text{ J}$ Therefore, $-P_{\text{ext}}\mathrm{\Delta }V=10.00\text{ J}$ Therefore, $P_{\text{ext}}\mathrm{\Delta }V=-10.00\text{ J}$ From formula (2), $\mathrm{\Delta }H=\mathrm{\Delta }U+P_{\text{ext}}\mathrm{\Delta }V$ Therefore, $\mathrm{\Delta }U=\mathrm{\Delta }H-P_{\text{ext}}\mathrm{\Delta }V$ $=-620\text{ J}-(-10.00\text{ J})=-610\text{ J}$ Pressure-volume work ($W$ = +10.00 J) and $\mathrm{\Delta }U$ = -610 J.