- #1
krootox217
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Homework Statement
The heat capacity of the solid NaCl from 500 K to 1074 K is given by [52.996 J*K-1*mol-1 – (7.86*10-3J*K-2*mol-1)*T + (1.97*10-5J*K-3*mol-1)*T2 ] and that of liquid NaCl from 1074 K to 1500 K is given by [125.637 J*K-1mol-1 – (8.187*10-2 J*K-2*mol-1)*T + (2.85*10-5 J*K-3*mol-1)*T2 ]. If Δfus HΘT, m = 28.158 kJ*mol-1 at 1074K, Please determine HΘ1500K - HΘ500K for NaCl.
Homework Equations
The Attempt at a Solution
I tried to solve this task, but I don't understand what they exactly want me to do.
Do I have to calculate the ΔH from 500K to 1074K with the given heat capacities? And what is ΔfusHΘ?