Question

Consider the following reaction and use the information given here to determine the value of $ \Delta {{S}_{surr}} $ at $ 398K $ . Also predict whether or not the given reaction will be spontaneous at this temperature.
 $ 4N{{H}_{3}}(g)+3{{O}_{2}}(g)\to 2{{N}_{2}}(g)+6{{H}_{2}}O(g)\ \ \ \ \Delta H=-1267kJ $
A. $ \Delta {{S}_{surr}}=+12.67\ kJ{{K}^{-1}} $ , reaction not spontaneous
B. $ \Delta {{S}_{surr}}=-12.67\ kJ{{K}^{-1}} $ , reaction spontaneous
C. $ \Delta {{S}_{surr}}=+50.4\ kJ{{K}^{-1}} $ , reaction not spontaneous
D. $ \Delta {{S}_{surr}}=+3.18\ kJ{{K}^{-1}} $ , reaction spontaneous
E. $ \Delta {{S}_{surr}}=-3.18\ kJ{{K}^{-1}} $ , it is not possible to predict the spontaneity of this reaction without more information.

Answer 1

Hint: For the given question, first we will calculate the change in entropy for surrounding by substituting the given values in its formula and then we can predict the spontaneity of the reaction on the basis of the value of free energy change which is calculated with the help of entropy and enthalpy of the reaction.
Formula used-
\[\Delta {{S}_{surr}}=-\dfrac{\Delta H}{T}\ \ \ \ \ \ ...(1)\]
 $ \Delta G=\Delta H-T\Delta S\ \ \ \ ...(2) $
Where, $ \Delta S $ is the change in entropy, $ \Delta G $ is the change in free energy, $ \Delta H $ is the enthalpy of the reaction and T is the temperature.

Complete answer:
The change in entropy can be defined as the change in the state of disorder of a thermodynamic system which is associated with the conversion of enthalpy or heat into work. It is a state function i.e., its value does not depend on the path of the thermodynamic process.
As per question, the given reaction is as follows:
 $ 4N{{H}_{3}}(g)+3{{O}_{2}}(g)\to 2{{N}_{2}}(g)+6{{H}_{2}}O(g)\ \ \ \ \Delta H=-1267kJ $
We need to calculate change in entropy at $ 398K $ for the given reaction. So, according to equation (1):
\[\Delta {{S}_{surr}}=-\dfrac{\Delta H}{T}\]
Substituting given values:
 $ \Rightarrow \Delta {{S}_{surr}}=-\dfrac{-1267}{398} $
 $ \Rightarrow \Delta {{S}_{surr}}=+3.18kJ{{K}^{-1}} $
Now, we know that a spontaneous reaction is a chemical reaction in which the formation of products is favoured at the conditions under which the reaction is occurring. We can use Gibbs free energy to determine the spontaneity of a chemical reaction. Criteria of spontaneity of a reaction on the basis of Gibbs free energy is as follows:
1. If $ \Delta G $ is negative, then the process is spontaneous.
2. If $ \Delta G $ is positive, then the process is nonspontaneous.
3. If $ \Delta G $ is zero, then the process is in equilibrium.
So, according to equation (2), the free energy can be expressed in terms of enthalpy and entropy as follows:
 $ \Delta G=\Delta H-T\Delta S $
Substituting values:
 $ \Rightarrow \Delta G=-1267-398\times 3.18 $
 $ \Rightarrow \Delta G=-2532.64kJ $
As the value of free energy is negative, the given reaction is spontaneous. Hence, the correct answer is option (D) i.e., $ \Delta {{S}_{surr}}=+3.18\ kJ{{K}^{-1}} $ and the reaction is spontaneous.

Note:
It is important to note that we can also predict the spontaneity of the given chemical reaction on the basis of calculated value of change in entropy as per second law of thermodynamics which states that the total entropy of the system either increases or remain constant for any process to be spontaneous that means $ \Delta {{S}_{total}}>0 $ .