Calculate the heat energy released when 28,9g of liquid mercury at 25.00° C is c

Question

Calculate the heat energy released when 28,9g of liquid mercury at 25.00°C is converted to solid mercury at its melting point.

constants for mercury at 1 atm

heat capacity of Hg(l)- 28.0 J/(mol×K)

melting point- 234.32 K

enthalpy of fusion- 2.29 kJ/mol

- ________ kJ

2) At 1 atm, how much energy is required to heat 71.0g of H2O(s) at -12.0°C to H2O(g) at 171.0°C?

- ______kJ

3) Two 20.0-g ice cubes at -10.0°C are placed into 235g of water at 25.0°C. Assuming no energy is transferred to or from the surroundings, calculate the final temperature of the water after all the ice melt.

heat capacity of H2O(s)- 37.7 J/(mol × K)

heat capacity of H2O(l)- 75.3 J/(mol × K)

enthalpy of fusion H2O- 6.01 kJ/mol

- _________°C

4) A certain substance has a heat of vaporization of 53.06 kJ/mol. At what Kelvin temperature will the vapor pressure be 6.50 times higher than it was at 287 K?

- ______K

Explanation / Answer

q1 = heat released moving T from 298 to m.p.
q1 = mol of Hg x heat capacity Hg x (Tfinal-Tinitial) = (28.9/200 ) x 28 x ( 298 - 234.32) = 257.6 J

q2 = heat released by freezing
q2 = mol Hg x heat fusion = (28.9/ 200) x 2.29 x 10^3 = 330.905 J

Total heat released = q1 + q2 = 257.6 + 330.905 = 588.505 J.

2) ice @ –12.0 °C --> ice @ 0.0 °C , q1 = mcT
ice @ 0.0 °C --> water @ 0.0 °C , q2 = mHf
water @ 0.0 °C --> water @ 100 °C q3 = mcT
water @ 100.0 °C --> steam @ 100.0 °C q4 = mHv
steam @ 100.0 °C --> steam @ 171.0 °C q5 = mcT

Total heat required = q(T) = q1 + q2 + q3 + q4 + q5
q(T) = mcT + mHf + mcT + mHv + mcT
= 71 ( 2.01 x 12 + 334 + 4.184 x 100 + 2257 + 2.010 x 71 ) = 2.255 x 10^ 5 J
c(ice) = 2.010 J/g°C
c(water) = 4.184 J/g°C
c(steam) = 2.010 J/g°C
Hf = 334 J/g
Hv = 2257 J/g

3)40.0 g ice x (1 mole ice / 18.0 g) = 2.22 moles ice
235 g H2O(l) x (1 mole H2O(l) / 18.0 g) = 13.1 moles H2O(l)

Heat lost by H2O(l) = (mass H2O(l))(heat capacity H2O(l))(Ti - Tf) = (13.1 moles)(75.3 J/mol K)(25.0 °C - Tf)
= 24660 - 986 Tf

The heat absorbed by the ice will take place in three steps: (1) warming the ice from -10.0 °C to 0.0 °C, (2) the actual melting of the ice to convert it into H2O(l), and (3) the warming of this new H2O(l).

Heat gained by ice = (mass ice)(heat capacity ice)(Tf - Ti) + (mass ice)(heat of fusion of ice) + (mass new H2O(l)(heat capacity H2O(l)(Tf - Ti)
= (2.22 moles)(37.7 J/mol K)(0.0 °C - (-10.0 °C)) + (2.22 moles)(6010 J/mol) + (2.22 moles)(75.3 J/mol K)(Tf - 0.0 °C) = 836.92 J + 13,300 J + 167 Tf

Heat lost by H2O(l) = heat gained by ice
836.92 + 13,300 + 167 Tf = 24,660 - 986 Tf
1153 Tf = 10523
Tf = 10523/ 1153 = 9.12 °C

4)Using Clausius Clapeyron equation.
ln(P/P) = (H/R)((1/T) - (1/T))
for vaporization:
P,P - the vapor pressures at absolute temperatures T and T respectively
H - enthalpy of vaporization
R - universal gas constant

P/P = 6.5
So solve Clausius Clapeyron equation for T:
ln(P/P) = (H/R)((1/T) - (1/T))
=>
T = 1/[ (1/T) - (R/H)ln(P/P) ]
= 1/[ (1/287 K) - (8.3145 Jmol¹K¹/ 53.06×10³ Jmol¹) ln(6.5) ]
= 313.37 K

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