2H2O(l)2H2(g) O2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 2.03 moles of H2O(l) react at standard conditions.

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

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answer

i think the answer is a) as air pressure in an area increases, the density of the gas particles in that area decreases.

hope it