Question
Bromine, \(Br_2 (l)\), and methanoic acid, HCOOH(aq), react in the presence of sulfuric acid.
\(Br_2 (l) + HCOOH(aq) → 2HBr(aq) + CO_2 (g)\)
(a) Suggest an experimental method that could be used to determine the rate of reaction.
(b) The sulfuric acid is a catalyst in this reaction. Explain how a catalyst increases the reaction rate.
(c) Methanoic acid can react with ethanol to produce an ester.
(i) Draw the full structural formula of the organic product and state its name.
Structural formula:
Name:
(ii) Predict the number of signals, and their splitting patterns, in the \(^1H\) NMR spectrum of this organic product.
Number of signals:
Splitting patterns:
(iii) State one reason why tetramethylsilane, TMS, is often chosen as an internal reference standard for the calibration of \(^1H\) NMR spectroscopy.
(d) (i) Write the equation for the complete combustion of ethanol.
(ii) Determine the enthalpy change for the combustion of ethanol, in kJ \(mol^{−1}\),using section 11 of the data booklet.
Answer/Explanation
Answer:
(a) «measure change in»
mass
OR
pressure
OR
volume of gas/\(CO_2\) produced
OR
«intensity of» colour
OR
«electrical» conductivity
OR
pH
with time
(b) provides an alternative reaction pathway AND lower activation energy/\(E_a\)
larger fraction/number of molecules with E ≥ \(E_a\)/enough energy «for a successful collision»
(c) (i) Structural formula:
Name:
ethyl methanoate
(ii) Number of signals:
3
Splitting patterns:
singlet/1
AND
quartet/4
AND
triplet/3
(iii) Any two of the following:
chemical shift/signal outside range of common chemical shift/signal
strong signal/12/all H atoms in same environment
singlet/no splitting of signal
volatile/easily separated/easily removed
inert/stable
soluble in most organic solvents
(d) (i) \(CH_3CH_2OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(g)\)
(ii) «bond breaking»
1 C-C + 5 C-H + 1 C-O + 1 O-H + 3 O=O / 346 + 5(414) + 358 + 463 +3(498) /
4731 «kJ»
«bond forming»
4 C=O + 6 O-H / 4(804) + 6(463) / 5994 «kJ»
DH «= 4731– 5994» = −1263 «kJ \(mol^{−1}\) »
Question
An equation for the combustion of propane is given below.
C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (g)
(a) Determine the standard enthalpy change, ΔHΘ, for this reaction, using section 11 of the data booklet. [3]
(b) Calculate the standard enthalpy change, ΔHΘ, for this reaction using section 12 of the data booklet. [2]
(c) Predict, giving a reason, whether the entropy change, ΔS Θ, for this reaction is negative or positive. [1]
(d) Calculate ΔS Θ for the reaction in J K–1, using section 12 of the data booklet.
The standard molar entropy for oxygen gas is 205 J K–1 mol–1. [2]
(e) Calculate the standard Gibbs free energy change, ΔGΘ, in kJ, for the reaction at 5 °C, using your answers to (b) and (d). Use section 1 of the data booklet.
(If you did not obtain an answer to (b) or (d) use values of –1952 kJ and +113 J K–1 respectively, although these are not the correct answers.) [2]
Answer/Explanation
Ans:
a
Bonds broken: 8(C–H) + 2(C–C) + 5(O=O) / 8 × 414 «kJ mol−1» + 2 × 346 «kJ mol−1» + 5 × 498 «kJ mol−1» / 6494 «kJ»
Bonds formed: 6(C=O) + 8(O–H) / 6 × 804 «kJ mol−1» + 8 × 463 «kJ mol−1» / 8528 «kJ»
«Enthalpy change = bonds broken − bonds formed = 6494 kJ − 8528 kJ =» −2034 «kJ»
b
4(−241.8 «kJ») AND 3(−393.5 «kJ») AND «1»(−105 «kJ»)
«∆HƟ = 4(−241.8 «kJ») + 3(−393.5 «kJ») – «1»(−105 «kJ») =» − 2043 «kJ»
c positive AND more moles «of gas» in products 1 3.
d
4 × 188.8 «J K −1» AND 3 × 213.8 «J K −1» AND «1 ×» 270 «J K −1» AND 5 × 205 «J K −1»
«∆S Ɵ = 4(188.8 J K −1 ) + 3(213.8 J K −1) – [1(270 JK −1) + 5(205 J K−1)] =» 102 «J K −1» 2 3. e «T = 5 + 273 =» 278 K «ΔGƟ = −2043 kJ − (278K × 0.102 kJ K −1) =» −2071 «kJ»