Answer:
The correct answer is -1946.89 j/K.
Explanation:
The reaction is:
2H2O (l) ⇔ 2H2 (g) + O2(g)
First there is a need to find dHfrxn = dHf(Products) - dHf(reactants)
With the help of standard thermodynamic table, the values of dHf can be obtained,
dHf(H2O) = -285.8 kJ/mol
dHf(H2) = 0 kJ/mol
dHf(O2) = 0 kJ/mol
dHrxn = 2 × dHf (H2) + dHf(O2) - 2 × dHf(H2O)
= 2 × 0 + 0-2 × (-285.8) kJ = +571.6 kJ
This value is for 2 moles of H2O, for 2.03 moles of H2O, the value will be,
dHrxn = +571.6 kJ/2 mol × 2.03 mol
= 580.174 kJ
The given temperature is 298 K.
So, the value of dSsurr = -dHrxn/T
= -580.174kJ/298 K
= -580174 j/298K
= -1946.89 j/K
