Question

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

Answer 1

The heat of vaporization can be calculated using the enthalpy changes of formation and the enthalpy changes of vaporization.

Step-by-step explanation:

The heat of vaporization of water at a given set of conditions can be calculated using the enthalpy changes of formation and the enthalpy changes of vaporization.

From the given information, we know that 241.8 kJ of heat is released when one mole of H2O forms from its elements, and 285.8 kJ is released when one mole of H2O is formed from its elements.

To find the ΔH of vaporization, we subtract the enthalpy change of formation of liquid water from the enthalpy change of formation of water vapor:

ΔHvap = ΔHformation (vapor) - ΔHformation (liquid) = 285.8 kJ/mol - 241.8 kJ/mol = 44 kJ/mol

Answer 2

44Kj

Step-by-step 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