Question

At a given set of conditions, 241.8 kJ of heat is released when 1 mol of H₂O(g) forms from its elements. Under the same conditions, 285.8 kJ is released when 1 mol of H₂O(l) forms from its elements. Find ΔH for the vaporization of water at these conditions.

Answer 1

The vaporization of water at given conditions is 44kJ.

What is Vaporization?

Vaporization is defined as transition of substance from liquid phase to vapor phase.

These are the equation for the reaction which is described in the question,

H2(g) + ½O2(g) ------- H2O(g) . ∆ H -241. 8 kJ ------- eqn (1)

H2(g) + ½o2(g) -------- H2o (l) .

∆ H = 285.8 kJ -------------eqn (2)

But from the eq (2) we can see that it moves from gas to liquid, we will re- write the equation for vaporization of water as

H2o(l) ------->> H2o(g) ---------- eqn (3)

But the equation from eq (2) the equation does go with vaporization so we can re- write as

H2o------- H2(g) + ½ o2(g)

∆H= 285.8 kj ------------ eq(4)

Now, ∆H of the Vaporization of water at these conditions , we will sum up eq(1) and eq(4)

∆H = 285.8 kj + (-241.8 kj ) = 44 kj

Thus, we calculated that the vaporization of water at given conditions is 44kJ.

learn more about ∆H :

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