Question

Consider the reaction 2SO2(g) O2(g) Right arrow. 2SO3(g). Substance Delta. Hf (kJ/mol) S (J/(molmc021-3. JpgK)) SO3(g) –396 130. 58 SO2(g) –297 191. 50 O2(g) 0 205. 00 What is the Delta. Grxn of this reaction, and would it be spontaneous or nonspontaneous at 300. 0 K? Use Delta. G = Delta. H – TDelta. S. –296 kJ/mol, spontaneous –296 kJ/mol, nonspontaneous –100 kJ/mol, spontaneous –100 kJ/mol, nonspontaneous.

Answer 1

The reaction has $\Delta G$ value of -296 kJ/mol, and the reaction is spontaneous. Hence, option A is correct.

The value of $\Delta G$ for the reaction is the change in the free energy. The $\Delta G$ for the reaction is calculated as:

$\Delta G=\Delta H-T\Delta S$

Computation for $\Delta G$ of the reaction

The given reaction is:
$\rm 2\;SO_2\;+\;O_2\;\to\;2\;SO_3$

The change in enthalpy and entropy is given by the difference in product and reactant.

Thus, the change in free energy is given as:

$\Delta G=(\Delta H_{product}-\Delta H_{reactant})-T(\Delta S_{product}-\Delta S_{reactant})\\\\\Delta G=(2(\Delta H_{SO_3})-(2(\Delta H_{SO_2})\;+\;\Delta S_{O_2}))-T((2(\Delta S_{SO_3})-(2(\Delta S_{SO_2}))$

Substituting the values in the above equation at a temperature of 300 K.

$\Delta G=(2(-396)-(2(-297)\;+\;(0)))-300\;(2(130.58)-(2(191.50)+(205)))\\\Delta G=(-792-(-1386))-300\;(261.16+588)\\\Delta G=-296\; \rm kJ/mol$

The negative value of $\Delta G$ represents the reaction to be spontaneous.

Thus, the reaction has $\Delta G$ value of -296 kJ/mol, and the reaction is spontaneous. Hence, option A is correct.

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