Calculate the standard internal energy change for $OF_{2(g)} + H_2O_{(g)} \longrightarrow 2 HF_{(g)} + O_{2(g)}$ at $300 \ K$,if $\Delta_{f} H^{\circ}$ of $OF_{2(g)}$,$H_2O_{(g)}$,and $HF_{(g)}$ are $20$,$-250$,and $-270 \ kJ \ mol^{-1}$ respectively. $[R = 8.314 \ J \ K^{-1} \ mol^{-1}]$ (in $kJ$)

For an ideal gas, consider only $P-V$ work in going from an initial state $X$ to the final state $Z$. The final state $Z$ can be reached by either of the two paths shown in the figure. Which of the following choice(s) is (are) correct? [take $\Delta S$ as change in entropy and $w$ as work done].
(A) $\Delta S_{X \to Z} = \Delta S_{X \to Y} + \Delta S_{Y \to Z}$
(B) $w_{X \to Z} = w_{X \to Y} + w_{Y \to Z}$
(C) $w_{X \to Y \to Z} = w_{X \to Y} + w_{Y \to Z}$
(D) $\Delta S_{X \to Y \to Z} = \Delta S_{X \to Y}$

For a dimerization reaction,$2 A_{(g)} \rightarrow A_{2(g)}$ at $298 \ K$,$\Delta U^{\ominus} = -20 \ kJ \ mol^{-1}$,$\Delta S^{\ominus} = -30 \ J \ K^{-1} \ mol^{-1}$,then the $\Delta G^{\ominus}$ will be........ $J$

$11.0 \ L$ of an ideal gas at a constant external pressure of $5 \ atm$ is compressed isothermally to a final volume of $1 \ L$. The heat absorbed and work done,respectively,during this compression (in $L \ atm$) are:

For the reaction $2X_{(g)} + Y_{(g)} \to 2Z_{(g)}$ at $298 \ K$,$\Delta U = -10.5 \ kJ$ and $\Delta S^o = -10.5 \ J/K$. Calculate $\Delta G^o$ for the reaction. Will the reaction be spontaneous or not? Explain.

Consider the reversible processes for $1.0 \ mol$ of an ideal gas. $w_1, w_2, w_3$ and $w_4$ represent work done (in calories) in the processes $1, 2, 3$ and $4$, respectively; $\Delta U_2$ and $\Delta U_4$ are changes in the internal energy for the processes $2$ and $4$, respectively. (use $R = 2 \ cal \ K^{-1} \ mol^{-1}$). The correct option is: