At a given set of conditions 241.8 kJ of heat is released when one mole of H2O forms from its elements. Under the same conditions 285.8 KJ is released when one mole of H2O is formed from its elements. Find Delta h of the vaporization of water at these conditions

44Kj

Explanation:

These are the equations for the reaction described in the question,

Vaporization which can be defined as transition of substance from liquid phase to vapor

H2(g)+ 1/2 O2(g) >H2O(g). Δ H

-241.8kj eqn(1)

H2(g)+ 1/2 O2(g) >H2O(l).

Δ H =285.8kj eqn(2)

But from the second equation we can see that it moves from gas to liquid, we we rewrite the equation for vaporization of water as

H2O(l) >>H2O(g)eqn(3)

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

H2O > H2(g)+ 1/2 O2(g)

Δ H= 285.8kj eqn(4)

To find Delta h of the vaporization of water at these conditions, we sum up eqn(1) and eqn(4)

Δ H=285.8kj +(-241.8kj)= 44kj

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

24

1 is the answer to this problem