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Calculate the work done by the expansion of hydrogen gas when 7.1 g of sodium react with water at constant temperature of 27°C and pressure of 1.7 atm. Use a precision of 2 significant digits. 2Na(s) + 2H2O(l) → 2NaOH(s) + H2(g)

Respuesta :

Answer:

-390J

Explanation:

The given reaction is,

2Na(s) + 2H2O —-> 2NaOH + H2

Number of moles of NaOH = Mass/molar mass

= 7.1/23 = 0.31 mol

Since, 2 mol Na produces 1 mol H2

Hence, 0.31 mol Na will produce

= 0.31 mol /2 = 0.155 mol H2

According to ideal gas equation,

PV = n RT

V = nRT/P

= 0.155mol x 0.0821 L atm K-1 mol-1 x (27+273 K) / 1.7atm

= 2.25L

Therefore,

Work of expansion = - PV = - 1.7 atm x 2.25L

= - 3.825atm L

= - 3.825 x 101.325 J = - 387.57 J

= - 390 J

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