Question

question 6
lead(ii) oxide can be reduced to form elemental lead through the reaction:
pbo(s) + c(s) → pb(s) + co(g)
calculate δh° for the above reaction given the following (all values are at 25°c):
(i) 2 pb(s) + o₂(g) → 2 pbo(s) δh° = -436.0 kj
(ii) c(s) + ½ o₂(g) → co(g) δh° = -110.5 kj

question 7
benzene (c₆h₆) undergoes combustion according to:
2 c₆h₆(l) + 15 o₂(g) → 12 co₂(g) + 6 h₂o(l) δh° = -6535 kj
all of the heat from the combustion of 39.26 ml of c₆h₆ is transferred to 7.280 l of water at 24.7°c. what will the final temperature of the water be, in °c?
(the density of c₆h₆(l) is 0.8765 g/ml. assume the density of water is 1.00 g/ml and its specific heat is 4.184 j/(g °c).)

Explanation:

Question 6

Step1: Reverse reaction (i)

Reverse reaction (i): \(2PbO(s)\to2Pb(s)+O_2(g)\), \(\Delta H^{\circ}= + 436.0\space kJ\)
Divide by 2: \(PbO(s)\to Pb(s)+\frac{1}{2}O_2(g)\), \(\Delta H^{\circ}=+218.0\space kJ\)

Step2: Use reaction (ii)

Reaction (ii): \(C(s)+\frac{1}{2}O_2(g)\to CO(g)\), \(\Delta H^{\circ}=- 110.5\space kJ\)

Step3: Add the two modified reactions

\(PbO(s)+C(s)\to Pb(s)+CO(g)\)
\(\Delta H^{\circ}=(+218.0)+(-110.5)\space kJ\)

Step1: Calculate mass of benzene

Mass of \(C_6H_6\), \(m =
ho\times V\). Given \(
ho = 0.8765\space g/mL\) and \(V = 39.26\space mL\)
\(m=0.8765\times39.26 = 34.41\space g\)

Step2: Calculate moles of benzene

Molar mass of \(C_6H_6,M=(6\times12 + 6\times1)=78\space g/mol\)
Moles of \(C_6H_6,n=\frac{m}{M}=\frac{34.41}{78}=0.4412\space mol\)

Step3: Calculate heat released by benzene combustion

For \(2\space mol\) of \(C_6H_6,\Delta H^{\circ}=-6535\space kJ\)
For \(0.4412\space mol\) of \(C_6H_6,q=\frac{0.4412}{2}\times(- 6535)=-1441\space kJ=-1441000\space J\) (heat released by benzene, heat absorbed by water \(q = 1441000\space J\))

Step4: Calculate mass of water

Mass of water, \(m_{water}=
ho_{water}\times V_{water}\). Given \(
ho_{water}=1.00\space g/mL\) and \(V_{water}=7280\space mL\)
\(m_{water}=7280\space g\)

Step5: Use heat - transfer formula \(q = mC\Delta T\)

\(q=m_{water}C_{water}(T - T_0)\). Given \(C_{water}=4.184\space J/(g^{\circ}C)\) and \(T_0 = 24.7^{\circ}C\)
\(1441000=7280\times4.184\times(T - 24.7)\)
\(T-24.7=\frac{1441000}{7280\times4.184}\)
\(T-24.7=\frac{1441000}{30460.32}\)
\(T-24.7 = 47.3\)
\(T=24.7 + 47.3\)

Answer:

\(107.5\)

Question 7