Q91 / 200 · NEET 2024 (Re-NEET) · Chemistry

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:

  • Correct answer
    A
    $\quad 19988.4 \mathrm{~J}$
  • B
    200 J
  • C
    1999 J
  • D
    1.9988 kJ

Solution

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

Hence, the correct answer is option: 'A'.

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