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Question: The molar entropies of \[H{I_{(g)}},{H_{\left( g \right)}},{I_{\left( g \right)}}\] at 298K are 206....

The molar entropies of HI(g),H(g),I(g)H{I_{(g)}},{H_{\left( g \right)}},{I_{\left( g \right)}} at 298K are 206.5, 114.6 and 180.73 mol−1k−1mo{l^{ - 1}}{k^{ - 1}} respectively. Using the ΔG\Delta G, given below, calculate the bond energy of HI.
HI(g)→H(g)+I(g):ΔG∘=271.8kJH{I_{\left( g \right)}} \to {H_{\left( g \right)}} + {I_{\left( g \right)}}:\Delta {G^ \circ } = 271.8kJ
(A) 282.4
(B) 298.3
(C) 290.1
(D) 315.4

Explanation

Solution

The entropy content of one mole of pure material at a normal state of pressure and any temperature of interest is known as the standard molar entropy. ...... According to the third law of thermodynamics, the entropy of a pure crystalline structure can only be 0Jmol−1k−1Jmo{l^{ - 1}}{k^{ - 1}} at 0 K.

Complete answer: We know that the sum of the entropies of the products minus the sum of the entropies of the reactants equals the entropy change in a chemical reaction. The coefficients of each variable must be taken into account in the entropy calculation, just as they must be in other balanced equations calculations (the n, and m terms below imply that the coefficients must be accounted for):
ΔS0=∑nnS0(products)−∑mmS0\Delta {S^0} = \sum\limits_n {n{S^0}(products) - \sum\limits_m {m{S^0}} } (reactants)
By substituting the values in the equation,
ΔS0=−206.5+114.6+180.7\Delta {S^0} = - 206.5 + 114.6 + 180.7
=88.8= 88.8
The equation ΔG0=ΔH0−TΔS0\Delta {G^0} = \Delta {H^0} - T\Delta {S^0}gives the free energy change accompanying a process.
ΔG0=ΔH0−TΔS0\Delta {G^0} = \Delta {H^0} - T\Delta {S^0}(standard condition),
Where ∆H is the change in enthalpy, ∆S is the change in entropy and ∆H is the change in temperature.
Here we need to find the value ofΔH0\Delta {H^0},
By substituting the values in the equation we get,
ΔH0=271.8+298×88.8×10−3\Delta {H^0} = 271.8 + 298 \times 88.8 \times {10^{ - 3}}
ΔH0=298.3\Delta {H^0} = 298.3
Hence the correct option is (B)- 298.3

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