Question

Use the following information on Cr to determine the amounts of heat for the three heating steps required to convert 126.3 g of solid Cr at 1760°C into liquid Cr at 2060°C. mp = 1860°C bp = 2672°C Enter in kJ. Useful data: \Delta HΔ Hfus = 20.5 kJ/mol; \Delta HΔ Hvap = 339 kJ/mol; c(solid) 44.8 J/g°C; c(liquid) = 0.94 J/g°C

Answer 1

Answer : The amount of heat required is, 639.3 KJ

Solution :

The conversions involved in this process are :

$(1):Cr(s)(1760^oC)\rightarrow Cr(s)(1860^oC)\\\\(2):Cr(s)(1860^oC)\rightarrow Cr(l)(1860^oC)\\\\(3):Cr(l)(1860^oC)\rightarrow Cr(l)(2060^oC)$

Now we have to calculate the enthalpy change.

$\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]$

where,

$\Delta H$ = enthalpy change or heat required = ?

m = mass of Cr = 126.3 g

$c_{p,s}$ = specific heat of solid Cr = $44.8J/g^oC$

$c_{p,l}$ = specific heat of liquid Cr = $0.94J/g^oC$

n = number of moles of Cr = $\frac{\text{Mass of Cr}}{\text{Molar mass of Cr}}=\frac{126.3g}{52g/mole}=2.428mole$

$\Delta H_{fusion}$ = enthalpy change for fusion = 20.5 KJ/mole = 20500 J/mole

Now put all the given values in the above expression, we get

$\Delta H=[126.3g\times 44.8J/g^oC\times (1860-(1760))^oC]+2.428mole\times 20500J/mole+[126.3g\times 0.94J/g^oC\times (2060-1860)^oC]$

$\Delta H=639342.4J=639.3KJ$     (1 KJ = 1000 J)

Therefore, the amount of heat required is, 639.3 KJ