Question

Consider the following reaction:

2NO(g)+O2(g)→2NO2(g)
Estimate ΔG∘ for this reaction at each of the following temperatures and predict whether or not the reaction will be spontaneous. (Assume that ΔH∘ and ΔS∘ do not change too much within the give temperature range.) I need to find the temperature are 298K and 702K. For 298K It is simple because at standard temperature
ΔG∘ = DG(products)- DG(reactants).

Answer 1

A.
`\mathbf{\Delta G^0 = -72.6 \ kJ/mol}` ; as such the reaction is said to be spontaneous since the value of
`\mathbf{\Delta G^0 }` is negative.

B.
`\mathbf{\Delta G^0_(702 \ K) = -13.29 \ kJ/mol.K}}` and the reaction is spontaneous

Step-by-step explanation:

The equation for this chemical reaction is :

`2NO_((g)) +O_(2(g)) \to 2NO_(2(g))`

Using the following relation to calculate
`\Delta G^0`;

`\Delta G^0 = [2(\Delta G^0_{NO_(2(g))}] - [1(\Delta G^0_{O_(2(g))})+ 2(\Delta G^0_{NO_(g)})]`

At 298 K; the standard Gibbs Free Energy for the formation are as follows:

`\Delta G^0_{NO_(2(g))} = 51.2 \ kJ/mol`

`\Delta G^0_{O_(2(g))} = 0`

`\Delta G^0_{NO_(g)}= 87.6 \ kJ/mol`

Replacing them into the above equation;

`\Delta G^0 = [2(51.2 \ kJ/mol}] - [1(0)+ 2(87.6 \ kJ/mol})]`

`\Delta G^0 = [102.4 \ kJ/mol}] - [175.2 \ kJ/mol})]`

`\mathbf{\Delta G^0 = -72.6 \ kJ/mol}`

Thus;
`\mathbf{\Delta G^0 = -72.6 \ kJ/mol}` ; as such the reaction is said to be spontaneous since the value of
`\mathbf{\Delta G^0 }` is negative.

B.

Using the same above chemical equation;

The relation used for calculating
`\mathbf{\Delta G^0}` of the reaction when the temperature is 702 K is:

`\Delta G^0_(702 \ K) = \Delta H^0_(xn) - T \Delta S^0_(rxn)`

where;

`\Delta G^0_(702 \ K) =` Gibbs free energy of the reaction at 702 K

`\Delta H^0_(xn)` = standard enthalpy of the reaction = -116.2 kJ/mol

`\Delta S^0_(rxn)` = standard entropy of the reaction = -146.6 J/mol/K

Temperature T = 702 K

`\Delta G^0_(702 \ K) = -1162. \ kJ/mol - 702 \ K ( -146.6 \ J/mol. K ((1 \ kJ )/(1000 \ J))`

`\Delta G^0_(702 \ K) = -1162. \ kJ/mol - 702 \ K ( 0.1466 \ kJ/mol.K})`

`\Delta G^0_(702 \ K) = -13.2868 \ kJ/mol.K}`

`\mathbf{\Delta G^0_(702 \ K) = -13.29 \ kJ/mol.K}}`

Thus
`\mathbf{\Delta G^0_(702 \ K) = -13.29 \ kJ/mol.K}}` and the reaction is spontaneous