Question

Consider the reaction at constant pressure:

2 Na(s) + 2 H2O(l) → 2 NaOH(aq) + H2(g)        ΔE = -419 kJ

The reaction does 55 kJ on the surrounding.

Calculate the value of the ∆H (in kJ) that is produced by the reaction.

How much heat (in kJ) will be produced after the formation of 200.0 mL of hydrogen gas at STP?

Answer 1

1.

deltaE = - 419 kJ

Work done on surroundings = - 55 kJ

Therefore,

Enthalpy change = deltaE + work done

deltaH = - 419 - 55

deltaH = - 474 kJ

2.

V = 200.0 mL = 0.2000 L

At STP,

P = 1 atm

T = 273 K

R = 0.0821 L.atm / K.mol

Ideal gas equation,

p V = n R T

n = p V / ( R T )

n = 1 x 0.2000 / ( 0.0821 x 273 )

n = 0.00892 mol

From the balanced equation, 1 mol of H2 is being produced.

Heat produced for 1 mol of H2 gas = - 474 kJ

The,

Heat produced for 0.00892 mol of H2 = - 474 x 0.00892 = - 4.23 kJ