Calculate the entropy change in surrounding when 1-00 mol of displaystyle H-2-O-l- is formed under standard condition displaystyle bigtriangleup-f-H-ominus -286 KJ- mol-1- -

Question

Calculate the entropy change in surrounding when 1-00 mol of  displaystyle H-2-O-l- is formed under standard condition  displaystyle bigtriangleup-f-H-ominus -286 KJ- mol-1- -

Toppr Admin · Accepted Answer

Formation of water under standard conditions is an exothermic process. During the process heated is given out by the system to surrounding. Hence,
$q_{s u r r} + 286 k J / m o l$
The expression for the entropy change is $Δ S = \frac{q}{t}$
Substitute values in the above expression,
$Δ S = \frac{286 \times 1000}{298} = 959.7 J / K / m o l$ .

Calculate the entropy change in surrounding when 1.00 mol of H2O(l) is formed under standard condition △fH⊖=−286KJmol−1.

Formation of water under standard conditions is an exothermic process. During the process heated is given out by the system to surrounding. Hence, qsurr+286kJ/molThe expression for the entropy change is ΔS=qtSubstitute values in the above expression,ΔS=286×1000298=959.7J/K/mol.

Calculate the entropy change in surrounding when 1.00 mol of $H_{2} O (l)$ is formed under standard condition $△_{f} H^{⊖} = - 286 K J m o l^{- 1}$ .