An enthalpy diagram for a particular reaction is given in the figure. Is it possible to decide the spontaneity of a reaction from the given diagram? Explain.
(N/A) No,it is not possible to decide the spontaneity of a reaction solely from the enthalpy diagram.
Enthalpy change $(\Delta_rH)$ is only one of the contributing factors in determining the spontaneity of a reaction.
Another essential factor,the entropy factor $(\Delta S)$,must also be taken into consideration.
The overall spontaneity is determined by the Gibbs free energy change,given by the equation $\Delta G = \Delta H - T\Delta S$,where a reaction is spontaneous only if $\Delta G < 0$.
Enthalpy change $(\Delta_rH)$ is only one of the contributing factors in determining the spontaneity of a reaction.
Another essential factor,the entropy factor $(\Delta S)$,must also be taken into consideration.
The overall spontaneity is determined by the Gibbs free energy change,given by the equation $\Delta G = \Delta H - T\Delta S$,where a reaction is spontaneous only if $\Delta G < 0$.
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$A$ process has $\Delta H = 200 \ J \ mol^{-1}$ and $\Delta S = 40 \ J \ K^{-1} \ mol^{-1}$. Choose the minimum temperature above which the process will be spontaneous.
The spontaneous nature of a reaction is impossible if
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View SolutionAt $273 \ K$ the maximum work done when pressure on $10 \ g$ of hydrogen is reduced from $10 \ atm$ to $1 \ atm$ under isothermal,reversible conditions is (Assume the gas behaves ideally) $(R=8.3 \ J \ K^{-1} \ mol^{-1})$
Calculate the work done in the following reaction at $27^{\circ} C$.
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$4 SO_{2(g)} + 2 O_{2(g)} \rightarrow 4 SO_{3(g)}$
$(R = 8.314 \ J \ K^{-1} \ mol^{-1})$ (in $J$)
Calculate the maximum work done in expanding $16 \ g$ of oxygen $(O_2)$ at $300 \ K$ and occupying a volume of $5 \ dm^3$ isothermally until the volume becomes $25 \ dm^3$. (in $J$)
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