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For dissolution of Ammonium nitrate, $\mathrm{NH_4NO_3(s) \longrightarrow NH_4^+(aq.)+ NO_3^{-}(aq.)}$ the entropy change is $\mathrm{110\ J/mol.K}$ at $\mathrm{298\ K}$ and the enthalpy change of the system is   $\mathrm{29.8\ kJ/mol}$ . What is the entropy change of surrounding (in J/mol.K) and comment if the reaction is spontaneous or not?
Option: 1
$-100,$ spontaneous

Option: 2
$+100,$   non spontaneous

Option: 3
$+10,$    non-spontaneous

Option: 4
$+10,$   spontaneous

Answers (1)

best_answer

For calculating, $\mathrm{\Delta S_{surr}}$ , we have to consider the heat absorbed by surroundings is equal to the heat lost by the system.

$\mathrm{q_{surr}+ q_{system}=0}$

$\therefore \mathrm{q_{surr}=- q_{system}=- \Delta H_{sys}}$

Thus, the entropy change of surrounding can be calculated as $\mathrm{\Delta S_{surr}=- \frac{\Delta H_{sys}}{T}}$

$\therefore \mathrm{\Delta S_{surr}=-\frac{29.8\times 10^{3}}{298}=-100\; J/mol.K}$

Now, the spontaneity of the reaction is decided by the total entropy change of the process.

$\mathrm{\Delta S_{total}=\left ( \Delta S_{sys}+\Delta S_{surr} \right )}$

Now,

$\mathrm{\Delta S_{sys}=110\; J/mol.K}$

$\mathrm{\Delta S_{surr}=-100\; J/mol.K}$

$\therefore\mathrm{\Delta S_{total}=110-100=10\; J/mol.K}$

Total entropy change is positive so, this is a spontaneous reaction.

Entropy Change in the surrounding is (-100 J/mol-K).

Hence, option number (1) is correct

Posted by

jitender.kumar

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