What information about water is needed to calculate the enthalpy change for converting 1 molmol h2oh2o(gg) at 100 ∘c∘c to h2oh2o(ll) at 80 ∘c∘c?

The enthalpy change for the given process is -42.3 kJ/mol

Explanation:

The processes involved in the given problem are:  

Pressure is taken as constant.

To calculate the amount of heat released at same temperature, we use the equation:

      ......(1)

where,

= enthalpy change

= latent heat of vaporization

To calculate the amount of heat absorbed at different temperature, we use the equation:

       .......(2)

where,

= enthalpy change

= specific heat capacity of medium

= final temperature

= initial temperature

Calculating the enthalpy for each process:

We are given:

Putting values in equation 1, we get:

We are given:

Putting values in equation 2, we get:

Enthalpy change of the reaction =

Enthalpy change of the reaction =

Hence, the enthalpy change for the given process is -42.3 kJ/mol

the reaction mixture contains some ammonia, plus a lot of unreacted nitrogen and hydrogen. the mixture is cooled and compressed, causing the ammonia gas to condense into a liquid. the liquefied ammonia is separated and removed. the unreacted nitrogen and hydrogen are then recycled back into the reactor .

the movement of aircraft through clouds can even cause lightning strikes. a single lightning event is a "flash", which is a complex, multi-stage explosively ejected.

explanation:

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