1.How many kilojoules of heat must be supplied to 6.2 g of ice at -35 o C to con
ID: 886984 • Letter: 1
Question
1.How many kilojoules of heat must be supplied to 6.2 g of ice at -35 oC to convert it to steam at 152 oC? Use the following specific heats in J/g-oC:
ice=2.0
liquid = 4.2
steam = 2.0
Water has the following phase change enthalpies in kJ/mol
heat of fusion at 273 K = 6.01
heat of vaporization at 373 K = 40.7
Express your answer to the nearest 0.1 kJ.
????? k/j
The heat of formation of liquid hydrazine (N2H4) is 50.63 kJ/mol. Use tabulated data for other heats of formation.
The tolerance on each question is only 0.03 kJ, so express all answers to 0.01 kJ.
H = kJ q = kJ w = kJ E = kJExplanation / Answer
1. q = m*sice*DT1 + n*DHf + m*swater*DT2 + n*DHvap + m*swater*DT3
m= mass of ice = 6.2 grams
S = specific heat of ice = 2.0 j/g.c , water = 4.2 j/g.c . , steam = 2.0 j/g.c.
DT1 = (0-(-35)) = 35 C
DHfus = 6.01 kj/mol , DHvap = 40.7 kj/mol.
n = w/mwt = 6.2/18 = 0.34 mole.
DT2 = (100 -0) = 100
DT3 = (152-100) = 52
q = (6.2*2.0*35) + (0.34*6.01*1000) + (6.2*4.2*100)+(0.34*40.7*1000)+(6.2*2.0*52)
= 18919.4 joule
= 19.5642 kj
2.
N2O4(g) + 2N2H4(l) ---> 3N2(g) + 4H2O(l).
DHrxn = (4*H2o + 3*n2 )- (2*n2h4+ 1*n2o4)
=(4*(-285.3) +3*0) - (2*50.6 + 9.2)
DHrxn = -1251.6 kj
no of moles of N2O4 = 7.91/92.011 = 0.086 mole
1 mole n2o4 = -1251.6 kj
0.086 mole n2o4 = -107.6 kj
DHrxn = -107.6 kj
W= DnRT = (7-3)*8.314*298 = 9.91 kj
DH = DU + DnRT
Du = (-107.6)-9.91 = -117.51 kj
q = -DH= -107.6 kj
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.