Question:
\text{Match the following parameters with a description for spontaneity.}
\text{A. } \Delta_r H^{\circ} = +, \Delta_r S^{\circ} = -, \Delta_r G^{\circ} = + \text{ → Non-spontaneous at high temperature}
\text{B. } \Delta_r H^{\circ} = -, \Delta_r S^{\circ} = -, \Delta_r G^{\circ} = \text{+ at high T} \text{ → Spontaneous at all temperatures}
\text{C. } \Delta_r H^{\circ} = -, \Delta_r S^{\circ} = +, \Delta_r G^{\circ} = - \text{ → Non-spontaneous at all temperatures}
\text{Codes:}
\text{1. A-3, B-1, C-2}
\text{2. A-1, B-2, C-3}
\text{3. A-1, B-3, C-2}
\text{4. A-3, B-2, C-1}
Solution:
$\text{Hint: For a spontaneous reaction,}$ $\Delta\text{G value is less than zero.}$
$\text{A. When}$ $\Delta_r\text{G}^{\ominus}$ $\text{is positive, the reaction is non-spontaneous at all temperatures}$
$\text{B. When}$ $\Delta_r\text{G}^{\ominus}$ $\text{is positive at high temperature means the reaction is non-spontaneous at high temperature.}$
$\text{C. When}$ $\Delta_r\text{H}^{\ominus} = \text{negative means it favours,}$ $\Delta_r\text{S}^{\ominus} = \text{positive means it also favours.}$ $\Delta_r\text{G}^{\ominus} = \text{negative means reaction is spontaneous at all temperatures.}$
$\text{Now analyzing each parameter set:}$
$\text{A:}$ $\Delta_r\text{H}^{\ominus} = \text{positive},$ $\Delta_r\text{S}^{\ominus} = \text{negative},$ $\Delta_r\text{G}^{\ominus} = \text{positive}$
$\text{Since both enthalpy and entropy are unfavorable and}$ $\Delta_r\text{G}^{\ominus}$ $\text{is positive, this reaction is non-spontaneous at all temperatures. This matches description 3.}$
$\text{B:}$ $\Delta_r\text{H}^{\ominus} = \text{negative},$ $\Delta_r\text{S}^{\ominus} = \text{negative},$ $\Delta_r\text{G}^{\ominus} = \text{positive at high temperature}$
$\text{Negative enthalpy favors the reaction but negative entropy opposes it. At low temperatures, enthalpy dominates, but at high temperatures, entropy becomes more important. So this reaction is non-spontaneous at high temperature, which matches description 1.}$
$\text{C:}$ $\Delta_r\text{H}^{\ominus} = \text{negative},$ $\Delta_r\text{S}^{\ominus} = \text{positive},$ $\Delta_r\text{G}^{\ominus} = \text{negative}$
$\text{Both enthalpy and entropy favor the reaction, resulting in a negative}$ $\Delta_r\text{G}^{\ominus}$. $\text{This makes the reaction spontaneous at all temperatures, matching description 2.}$
$\text{Therefore, the correct mapping is:}$
$\text{A} \rightarrow 3 \text{ (Non-spontaneous at all temperatures)}$
$\text{B} \rightarrow 1 \text{ (Non-spontaneous at high temperature)}$
$\text{C} \rightarrow 2 \text{ (Spontaneous at all temperatures)}$
$\text{Looking at the codes, this corresponds to option 1.}$