Calculate the enthalpy change (in kJ) associated with the conversion of 25.0 gra

Question

Calculate the enthalpy change (in kJ) associated with the conversion of 25.0 grams of ice at -4.00degree C to water vapor at 106.0degree C. The specific heats of ice, water, and steam are 2.09 J/g-K. 4.18 J/g-K. and 1.84 J/g-K, respectively. For H_2O, Delta H_fas = 6.01 kJ/mol and Delta Hvap = 40.67 kJ/mol. 64.8 75.8 11000 12000 112 Ethanol (C_2H_5OH) melts at -114degree C. The enthalpy of fusion is 5.02 kJ/mol. The specific heats of solid and liquid ethanol are 0.97 J/g-K and 2.3 J/g-K, respectively. How much heat (kJ) is needed to convert 25.0 g of solid ethanol at -135degree C to liquid ethanol at -50degree C? 207.3 -12.7 6.91 4192 9.21

Explanation / Answer

We require 2 type of heat, latent heat and sensible heat

Sensible heat (CP): heat change due to Temperature difference

Latent heat (LH): Heat involved in changing phases (no change of T)

Then

Q1 = m*Cp ice * (Tf – T1)

Q2 = m*LH ice

Q3 = m*Cp wáter * (Tb – Tf)

Q4 = m*LH vap

Q5 = m*Cp vap* (T2 – Tb)

Note that Tf = 0°C; Tb = 100°C, LH ice = 334 kJ/kg; LH water = 2264.76 kJ/kg.

Cp ice = 2.01 J/g°C ; Cp water = 4.184 J/g°C; Cp vapor = 2.030 kJ/kg°C

Then

Q1 = 25*2.01 * (0 – -4) = 201

Q2 = 25*334 = 8350

Q3 = 25*4.184 * (100 – 0)

Q4 = 25*2264.76

Q5 = 25*2.03* (106– 100)

QT = Q1+Q2+Q3+Q4+Q5 = 201 + 8350+10460+56619+304.5 = 75934.5 J = 75.9 kJ

choose B since it is the nearest answer

Q.

Qsolid = 25*0.97* (-114 – - 135) = 509.25

Qmelting = 25/46*5.02*1000= 2728.260

Qliquid = 25*2.3* (-50 --114) = 3680

Qtotal = 3680+2728.260+509.25= 6917.51 = 6.92 kJ

choose C

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