Calculate the work expansion when 100 gram of water is electrolysed at a constant pressure of 1 ATM and temperature of 25 degree Celsius

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Solution

Dear Student,

The reaction scheme is,

$2 H_{2} O \to O_{2} + 2 H_{2} m o l e s o f H_{2} O = \frac{100 g m}{18 \frac{g m}{m o l}} = 5.55 m o l m o l e s o f H_{2} = 5.55 m o l e s o f O_{2} = 2.775 m o l W_{H_{2}} = - n R T = - 5.55 \times 8.3145 \times 298 = - 13.75 k J W_{O_{2}} = - n R T = - 2.775 \times 8.3145 \times 298 = - 6.876 k J W_{t o t a l} = W_{H_{2}} + W_{O_{2}} = (- 13.75) + (- 6.876) = - 20.626 k J$

Regards.

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Calculate the work expansion when 100 gram of water is electrolysed at a constant pressure of 1 ATM and temperature of 25 degree Celsius

Dear Student, The reaction scheme is, 2H2O→O2+2H2moles of H2O=100 gm18 gmmol=5.55 molmoles of H2=5.55moles of O2=2.775 molWH2=-nRT=-5.55×8.3145×298=-13.75 kJWO2=-nRT=-2.775×8.3145×298=-6.876 kJWtotal=WH2+WO2=(-13.75)+(-6.876)=-20.626 kJ Regards.