Question

Consider the following reaction at constant P. Use the information here to determine the value of ΔSsurr at 355 K. Predict whether or not this reaction will be spontaneous at this temperature. 2NO(g) + O2(g) → 2NO2(g) ΔH = -114 kJ

Answer 1

`\Delta S_(surr) = + 0.32113\: kJ/K`

Step-by-step explanation:

Given: Entropy of surrounding: ΔSsurr = ?

Temperature: T= 355 K

The change in enthalpy of reaction: ΔH = -114 kJ

Pressure: P = constant

As we know, ΔH = -114 kJ ⇒ negative

Therefore, the given reaction is an exothermic reaction

Therefore, Entropy of surrounding at constant pressure is given by,

`\Delta S_(surr) = (-\Delta H)/(T)`

`\therefore \Delta S_(surr) = -\left ((-114 kJ)/(355 K) \right ) = + 0.32113\: kJ/K > 0`

In the given reaction:

2NO(g) + O₂(g) → 2NO₂(g)

As, the number of moles of gaseous products is less than the number of moles of gaseous reactants.

`\therefore \Delta S_(system) < 0`

As we know, for a spontaneous process, that the total entropy should be positive.

`\Delta S_(total) = \Delta S_(surr) + \Delta S_(system) > 0`

Therefore, at the given temperature,

• if
  `\Delta S_(surr) > \Delta S_(system) \Rightarrow \Delta S_(total) > 0` then the given reaction is spontaneous

• if
  `\Delta S_(surr) < \Delta S_(system) \Rightarrow \Delta S_(total) < 0` then the given reaction is non-spontaneous