Question

6) If 15.4 g piece of aluminum is dropped into 654 g of water at 264 °C. The initial temperature of aluminum is 98.4°C. Calculate the final temperature of both water and aluminurn? (Specific heat of water and aluminum is from #1) 7) Calculate Enthalpy of the following reaction (Hz) from standard enthalpies of formation (H) from the table below. (a) C2H5OH(L) + 3 029) - 2 CO2(g) + 3 H2O(1) (b) 3 N0369) + H20(1) - 2 HNO3(aq) + NO(g)

Answer 1

6)

mass of Aluminum = 15.4 g

mass of Water = 65.4 g

specific heat of Al = 0.901 J/gC

specific heat of water = 4.184J/gC

Initial temperature of water = T1= 26.4C

Initial temperature of Al = 98.4C

Heat lost of by Al =Heat gained by Water

Q of heat = Q of water

(ms$\Delta$T) of Al = (ms$\Delta$T) of water

15.4 x 0.901 x (98.4 - T2) = 65.4 x 4.184 x ( T2 - 26.4)

1365.34 - 13.875 T2 = 273.634T2 - 7223.927

287.509 T2 = 8589.267

T2= 29.87C

Final temperature of the system = 29.87C

7)

a) C2H5OH(l) + 3 O2(g) ----------------- 2CO2(g) + 3 H2O(l)

$\Delta$Hrxn = 2 x $\Delta$ Hf of CO2 + 3 x $\Delta$ Hf of H2O - [ $\Delta$ Hf of C2H5OH + 3 x $\Delta$ Hf of O2]

$\Delta$Hrxn =[ 2 x( -393.51) + 3 x ( - 285.83) ] - [ - 276.98 + 3 x 0.0]

$\Delta$Hrxn = - 1367.53 KJ

b)

3 NO2(g) + H2O(l) ----------------- 2 HNO3(aq) + NO(g)

$\Delta$Hrxn = 2 x $\Delta$ Hf of HNO3 + $\Delta$ Hf of NO - [3x $\Delta$ Hf of NO2 + $\Delta$ Hf of H2O]

$\Delta$Hrxn = 2 x ( - 207.4) + 90.25 - [ 3 x 33.18 + ( - 285.83)]

$\Delta$Hrxn = - 138.26 KJ.