For the change $H_2O_{(l)} \to H_2O_{(g)}$ at $P = 1 \ atm$ and $T = 373 \ K$,the free energy change $\Delta G = 0$. This indicates that:
- A$H_2O_{(l)}$ is in equilibrium with $H_2O_{(g)}$
- BWater boils spontaneously at $373 \ K$
- CWater does not boil spontaneously at $373 \ K$
- DCondensation of water vapour occurs spontaneously at $373 \ K$
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Similar Questions
In which of the following conditions will a reaction always be spontaneous at all temperatures?
Medium
View SolutionThe entropy and enthalpy changes for the reaction $CO_{(g)} + H_2O_{(g)} \rightleftharpoons CO_{2(g)} + H_{2(g)}$ at $300 \ K$ and $1 \ atm$ are respectively $-42.4 \ J \ K^{-1}$ and $-41.2 \ kJ$. The temperature at which the reaction will go in the reverse direction is (in $K$)
Calculate the value of $\Delta G$ for the following reaction at $300 \ K$.
$H_2O_{(s)} \longrightarrow H_2O_{(l)}$
$(\Delta H = 7 \ kJ, \Delta S = 24.8 \ J \ K^{-1})$
$H_2O_{(s)} \longrightarrow H_2O_{(l)}$
$(\Delta H = 7 \ kJ, \Delta S = 24.8 \ J \ K^{-1})$
Which of the following processes will never be a spontaneous process?
Easy
View SolutionAt a temperature of $25^\circ C$,if $\Delta H < T \Delta S$,then $\Delta G$ = ?
Easy
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