Standard enthalpy and standard entropy for the oxidation of $N H_{3}$ at $298$ K are $- 382.64$ $kJ {mol}^{- 1}$ and $- 145.6$ $J {mol}^{- 1} K^{- 1}$ respectively. Standard free energy change (in ${kJ mol}^{- 1}$ ) for the same reaction at $298$ K is:

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If the standard internal energy change for the reaction OF2(g)+H2O(g)⟶O2(g)+2HF(g), at 298 K is x kJ. [Given standard enthalpies of formation in kJmol−1 are OF2(g)=+20;H2O(g)=−250andHF(g)=−270.]Then −100x is

ΔHorxn=ΔHoO2+2ΔHoHF−[ΔHoOF2+ΔHoH2O]ΔHorxn=0+2(−270)−[+20−250]=−310kJ=−310000J ΔEorxn=ΔHorxn−ΔnRT ΔEorxn=−310000−(1×8.314×298)=−312477J=−312.47kJ −100x=31247

If the standard internal energy change for the reaction $O F_{2} (g) + H_{2} O (g) ⟶ O_{2} (g) + 2 H F (g)$ , at 298 K is $x$ $k J$ .
[Given standard enthalpies of formation in $k J m o l^{- 1}$ are $O F_{2} (g) = + 20; H_{2} O (g) = - 250 a n d H F (g) = - 270$ .]
Then $- 100 x$ is