Predict the value of \(\Delta H_{\mathrm{f}}^{\circ}\) (greater than, less than, or equal to zero) for these elements at \(25^{\circ} \mathrm{C}\) : (a) \(\mathrm{Br}_{2}(g)\) and \(\mathrm{Br}_{2}(l)\) (b) \(\mathrm{I}_{2}(g)\) and \(\mathrm{I}_{2}(s)\)

Short Answer

Expert verified
The \(\Delta H_{\mathrm{f}}^{\circ}\) for \(Br_{2}(g)\) and \(I_{2}(g)\) is greater than zero, while for \(Br_{2}(l)\) and \(I_{2}(s)\) the \(\Delta H_{\mathrm{f}}^{\circ}\) is equal to zero.

Step by step solution

01

Evaluate Bromine

First consider bromine(\(Br_2\)). The standard state of bromine at \(25^{\circ} \mathrm{C}\) is liquid. Thus, for the formation of bromine in the gaseous state, \(\Delta H_{\mathrm{f}}^{\circ}\) would be greater than zero as energy is needed to convert it from the standard state to gaseous state. On the other hand, for the formation of bromine in the liquid state, which is its standard state, \(\Delta H_{\mathrm{f}}^{\circ}\) would be equal to zero, because there is no change needed for it to reach the standard state.
02

Evaluate Iodine

Next, consider iodine(\(I_2\)). The standard state of iodine at \(25^{\circ} \mathrm{C}\) is in its solid state. Therefore, for the formation of iodine in the gaseous state, \(\Delta H_{\mathrm{f}}^{\circ}\) would be greater than zero, because energy is required to convert it from the standard state to gaseous state. In contrast, for the formation of iodine in the solid state, which is its standard state, \(\Delta H_{\mathrm{f}}^{\circ}\) would be equal to zero, as there's no change needed for it to reach the standard state.

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