The enthalpy change H for the reaction- N2g- 3H2g- 2NH3g is -92-38kJ at 298 K- The internal energy change U at 298 K is-

If at $298 K$ the bond energies of C - H, C - C, C - C and H - H bonds are respectively 414, 347, 615 and 435 $k J m o l^{- 1}$ , the value of enthalpy change for the reaction $H_{2} C = C H_{2} (g) \to H_{3} C - C H_{3} (g)$ at 298K will be

Q. What will be the standard internal energy change for the reaction at 298 K?

O F_{2 (g)} + H_{2} O_{(g)} \to O_{2 (g)} + 2 H F_{(g)}; △ H^{0}

= - 310 kJ.

If at 298 K the bond energies of C - H, C - C, C=C and H - H bonds are respectively 414, 347, 615 and 435 kJ $m o l^{- 1}$ , the value of enthalpy change for the reaction

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The enthalpy change (△H) for the reaction, N2(g)+3H2(g)→2NH3(g) is -92.38kJ at 298 K. The internal energy change △U at 298 K is:

The correct option is A -87.42 kJFor isothermal process,△E=△H−△nRT=−92.38−(−2×8.31×2981000)=−92.38+4.95=−87.43kJHence, option B is correct.

The enthalpy change $(△ H)$ for the reaction, $N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g)$ is -92.38kJ at 298 K. The internal energy change $△ U$ at 298 K is:

The correct option is A -87.42 kJ
For isothermal process,

$△ E = △ H - △ n R T$

$= - 92.38 - (\frac{- 2 \times 8.31 \times 298}{1000})$

$= - 92.38 + 4.95 = - 87.43 k J$

Hence, option B is correct.