Question

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

Answer 1

Hint: 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 0\[Jmo{l^{ - 1}}{k^{ - 1}}\] 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):
\[\Delta {S^0} = \sum\limits_n {n{S^0}(products) - \sum\limits_m {m{S^0}} } \](reactants)
By substituting the values in the equation,
\[\Delta {S^0} = - 206.5 + 114.6 + 180.7\]
\[ = 88.8\]
The equation \[\Delta {G^0} = \Delta {H^0} - T\Delta {S^0}\]gives the free energy change accompanying a process.
\[\Delta {G^0} = \Delta {H^0} - T\Delta {S^0}\](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 of\[\Delta {H^0}\],
By substituting the values in the equation we get,
\[\Delta {H^0} = 271.8 + 298 \times 88.8 \times {10^{ - 3}}\]
\[\Delta {H^0} = 298.3\]
Hence the correct option is (B)- 298.3

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