At the normal melting point of nacl, 801 degrees c, its enthalpy of fusion is 28.8 kj / mol. the density of the solid is 2.165 g/cm^3 , and the density of the liquid is 1.733 g/ cm^3. what pressure increase is needed to raise the melting point by 1.00 degrees c?

The increase in pressure is 0.003 atm

Explanation:

To calculate the final pressure, we use the Clausius-Clayperon equation, which is:

where,

= initial pressure which is the pressure at normal boiling point = 1 atm

= final pressure = ?

= Enthalpy change of the reaction = 28.8 kJ/mol = 28800 J/mol     (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

= initial temperature =

= final temperature =

Putting values in above equation, we get:

Change in pressure =

Hence, the increase in pressure is 0.003 atm

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.

explanation:

0.07 g/cm³

1.37 g/cm³