4. Consider the decomposition of HCl(g): 2HCl(g) H2(g) + Cl2(g) H = 185 kJ (a) I
ID: 895961 • Letter: 4
Question
4. Consider the decomposition of HCl(g):
2HCl(g) H2(g) + Cl2(g) H = 185 kJ
(a) Is heat absorbed or released in this process? Answer: ______________________________________ (b) Calculate the value of H for the decomposition of 4 moles of HCl(g).
(c) Determine the value of H for the following reaction: H2(g) + Cl2(g) 2HCl(g)
(d) Calculate the value of H for the following reaction: 4H2(g) + 4Cl2(g) 8HCl(g)
5. (a) Calculate the specific heat of copper (in J/g·K) if 12.4 J of heat is required to raise the temperature of a 9.12 g sample of copper by 3.52 K.
(b) Calculate the amount of heat in Joule required to raise the temperature of a 145 g of copper by 12.4 K.
6. When 0.100 L of 2.00 M potassium hydroxide and 0.100 L of 2.00 M nitric acid are mixed in a coffee-cup calorimeter, the temperature increases from 19.18oC to 27.11oC. Calculate the enthalpy change for the neutralization reaction in kJ. Assume that the heat capacity of the calorimeter apparatus is 141 J/K, the specific heat of the solution is 4.18 J/g·K, the total volume of the solution is 0.200 L, and the density of the solution is 1.04 g/mL.
Explanation / Answer
4.
a) As DHrxn is +ve : heat absorbed
b) 2 mole = 185 kj
4 mole = 185*4/2 = 370 kj
c) H2(g) + Cl2(g) 2HCl(g)
its opposite to origibal one
DHrxn = -185 kj
d) 4H2(g) + 4Cl2(g) 8HCl(g)
DHrxn = 4*(-185) = -740 kj
5. q = m*s*DT
m = 9.12 , S = specificheat = ?
DT = 3.52 k q = 12.4 kj
12.4 = 9.12*s*3.52
s = 0.386 j/k.g
6. mass of solution = 0.2*1.04 = 0.208 grams
specific heat of the solution = 4.18 J/g·K
DT = 27.11-19.18 = 7.93
q = msDT = 0.208*4.18*7.93 = 6.9 j
No of moles of KOH= 0.1*2 = 0.2 mole
DH = 6.9/0.2 = 34.5 j/mol
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