For the following reaction at 300 K $\mathrm{A}_2(\mathrm{~g})+3 \mathrm{~B}_2(\mathrm{~g}) \rightarrow 2…

For the following reaction at 300 K $\mathrm{A}_2(\mathrm{~g})+3 \mathrm{~B}_2(\mathrm{~g}) \rightarrow 2 \mathrm{AB}_3(\mathrm{~g})$ the enthalpy change is +15 kJ then the internal energy change is:
  1. $\quad 19988.4 \mathrm{~J}$
  2. 200 J
  3. 1999 J
  4. 1.9988 kJ

Solution

$\mathrm{A}_2(\mathrm{~g})+3 \mathrm{~B}_2(\mathrm{~g}) \rightarrow 2 \mathrm{AB}_3(\mathrm{~g})$ $\Delta n_{(\mathrm{g})}=\mathrm{n}_{(\mathrm{P})}-\mathrm{n}_{(\mathrm{R})}$ $=2-3-1=-2$ $\Delta H=\Delta U+\Delta n_g R T$ $15 \times 1000=\Delta U-2 \times 8.314 \times 300$ $\Delta U=15000+600 \times 8.314$ $=15000+6 \times 831.4$ $=15000+4988.4$ $\Delta U=19988.4 \mathrm{~J}$

Asked in: NEET 2024 (Re-NEET)

Practice more Chemical Thermodynamics questions on Aicharya