The vaporization of water at given conditions is 44kJ.
<h3>What is Vaporization? </h3>
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 :
brainly.com/question/24170335
#SPJ4
