Chemical Energetics

Question 1

3.0 of Zinc powder (an excess) was added to 30.0 cm^3 of copper (||) sulphate solution of conc. of 1.0 mol dm^-3. The temp rise of the mixture was 47.6K (assuming the heat capacity of the solution is 4.2JK^-1 g^-1.
Calculate the enthalpy change of the reaction.

Answer 1

ΔH = -(30 x 4.2 x 47.6)/3 ✔️✔️

Question 2

C(s, graphite) + O2(g) —> CO2(g)
ΔH = -394 kJ mol^-1
CO(g) + 0.5O2(g) —> CO2(g)
ΔH = -283kJ mol^-1
These are the enthalpy change of combustion of graphite and of CO.

Calculate the enthalpy change for the reaction of graphite with oxygen to give carbon monoxide.

Answer 2

-394 + 283 = -111 kJ mol^-1 ✔️

Question 3

The molar enthalpy change of combustion of some alkanes is below in kJ mol^-1 
C3H8         -2219
C4H10       -2877
C5H12       -3509
C6H14       -4163

Another alkane was found to have an enthalpy change of combustion of -6125 kJ mol^-1.

Which alkane is it?

Answer 3

C9H20 ✔️

Question 4

In an experiment to measure the enthalpy change of a reaction involving gases, which of the following conditions must always be kept constant?

1. Pressure
2. Temperature
3. Volume
4. Temperature and pressure

Answer 4

1 ✔️

Question 5

Define the term enthalpy change of combustion.

Answer 5

The amount of energy used to combust the products in their natural state (in 1 mol) ✔️

Question 6

1.76g of pentan-1-ol was burnt. The energy was used to heat 250cm^3 of water from 24 to 78 degrees.

Calculate the energy released in kN during combustion is 1.76g of pentan-1-ol, CH3(CH2)4OH.

Answer 6

q=mcΔT

q=(250 x 4.18 x (78-24)= 56.43kJ
✔️✔️

Question 7

Calculate the amount (in mol) of pentan-1-ol when 1.76g is burned.

Answer 7

Mol = mass/Mr

   = 1.76/88 = 0.02mol ✔️

Question 8

What are standard conditions?

Answer 8

298K and 101KPa ✔️

Question 9

Write the state symbols:

6C(…) + 7H2(…) —> C6H14(…)

Answer 9

6C(s) + 7H2(g) —> C6H14(L) ✔️

Question 10

It is very difficult to determine the standard enthalpy change of formation of hexane directly. Suggest a reason why.

Answer 10

Because many products can be made so it’s hard to calculate directly one. ✔️

Question 11

Calculate the standard enthalpy change of formation of hexane using the standard enthalpy changes of combustion below:

Substance ΔHc。/KJmol^-1
C -394
H2 -286
C6H14 -4163

Answer 11

-4366 -(-394-286) = -203 kJmol-1 ✔️✔️

Question 12

Suggest two reasons why standard enthalpy changes or combustion determined experimentally are less exothermic than the calculated theoretical values.

Answer 12

1. Energy can escape

2. Combustion wasn’t completed ✔️✔️

Question 13

Write the equation for aerobic respiration of C6H12O6.

Answer 13

C6H12O6 + 6O2 —> 6CO2 + 6H2O ✔️✔️

Question 14

Suggest why the enthalpy change of formation of C6H12O6 cannot be determined directly.

Answer 14

Because activation energy was involved ✔️

Question 15

What’s the meaning of the term activation energy?

Answer 15

The minimum amount of energy to overcome to trigger the start of a reaction ✔️✔️

Question 16

H and Cl are reacted together to down HCl —
H2(g) + Cl2(g) -> 2HCl (g)
Calculate the bond enthalpy for the H-Cl bond using table below:

Bond Bond enthalpy
H-H +436
Cl-Cl +243

The ΔH = -184kJ Mol^-1

Answer 16

436 + 243 = 679

679 + 184 = 863/2 = 431 ✔️✔️