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SS2: CHEMISTRY - 2ND TERM

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  1. Water & Solution I | Week 1
    10 Topics
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    1 Quiz
  2. Water, Solution and Solubility | Week 2 & 3
    9 Topics
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  3. Air | Week 4
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  4. Pollution | Week 5
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  5. Rate of Chemical Reaction | Week 6 & 7
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  6. Energy and Energy Effect I | Week 8 & 9
    7 Topics
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  7. Energy and Energy Effect II | Week 10 & 11
    6 Topics
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  8. Chemical Equilibrium | Week 12
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SS2: CHEMISTRY – 2ND TERM Energy and Energy Effect I | Week 8 & 9 SS2: Chemistry Theory Questions – Energy and Energy Effect I
Lesson 6, Topic 7
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SS2: Chemistry Theory Questions – Energy and Energy Effect I

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Question 1:

[a]  Explain the terms exothermic and endothermic reactions. What is the sign of ∆H in each of the reactions?

[b] Draw an energy profile diagram for each of the following reactions.
i. H2[g] + I2[s] → 2HI[g]  ∆H = +53.6KJmol-1
ii. C[s] + O2[g] → CO2[g]  ∆H = -480KJmol-1

View Answers

Question 2:

[a]
[i]. Define enthalpy of combustion.
(ii)Why is the heat of combustion always negative
(iii)Name the instrument used for the determination of heat of combustion in the laboratory.
(iv)When 0.06 mole of methane is burnt in air, the heat liberated raises the temperature of 100g of water from 70°C to 85°C.
What is the heat of combustion of methane? [ specific heat capacity = 4.2Jg-1 °C-1]

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 Question 3:

Define the following terms
i. Standard heat of formation 
ii. Heat of solution
iii. Heat of neutralization
iv. Heat of combustion

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Question 4:

(a). Explain why the heat of neutralization between strong acid and strong alkali pair is of an approximate value of -57KJmol-1.
(b). when 1 mole of hydrochloric acid neutralized 1 mole of potassium hydroxide, -57.5KJ of heat was given off. Calculate the heat evolved when
i. 1.5 mole of hydrochloric acid
ii. 3 moles of hydrochloric acid react with excess potassium hydroxide solution.

View Answers

Question 5:

[a] The heat of formation of carbon[iv]oxide is -382KJ/mole. Calculate the heat given off when carbon is burnt to form 8g of carbon[iv]oxide gas. [C = 12, O = 16]

[b] The enthalpy change of a reaction is as follows:
CH4[g] + 2O2[g] → CO2[g] + 2H2O[g]   is -890.7KJ

Given that
∆Hθf [CO2] = -193.3KJmol-1
∆HθF [H2O] = -286.1KJmol-1
Calculate the heat of combustion of methane

[c] In an experiment, 2 g of methanol was burned completely in air.
i. Write an equation for the combustion of methanol
ii. Using your equation and the data given below, calculate the ∆cH (standard
enthalpy change of combustion) for methanol.
iii. Comment on whether this is an endothermic or exothermic reaction (using
your answer to (b))

Data:

The enthalpy changes of formation are as follows:

Substance Standard Enthalpy Change of Formation
(∆fH) / kJ mol–1
CH3OH(l)–234
CO2(g)–394
H2O(l)–286
View Answers

Question 6:

The enthalpy change of formation of NaCl was under investigation. 75 cm3 of 1 M
HCl was poured into a beaker then 75 cm3 NaOH 1 M was added to the HCl.
(a) Write an equation for the neutralisation of HCl by NaOH.
(b) Using your equation and the data given above, calculate the ∆rH of this
reaction
(c) Comment on the sign and magnitude of the value calculated in part (b).

Data:

The enthalpy changes of formation are as follows:

Substance Standard EnthalpyChange of Formation / kJ mol–1
H2O(l)–286
HCl(aq)–167
NaOH(aq)–425
NaCl(aq) –411
View Answers

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Question 1

[a]  Explain the terms exothermic and endothermic reactions. What is the sign of ∆H in each of the reactions?

Answer:

Exothermic reaction is one which releases energy as heat to the environment ∆H= -ve

Endothermic reaction is one which gains energy as heat from the surrounding ∆H= +ve


[b] Draw an energy profile diagram for each of the following reactions.

i. H2[g] + I2[s] → 2HI[g]  ∆H = +53.6KJmol-1

Answer:


ii. C[s] + O2[g] → CO2[g]  ∆H = -480KJmol-1

Answer:

 

Question 2

[a]
(i). Define enthalpy of combustion.

Answer: This is the amount of heat evolved or liberated when 1mole of a substance is burnt completely in excess oxygen under standard conditions

(ii) Why is the heat of combustion always negative

Answer:  Because its an exothermic reaction (Energy is lost)

(iii)Name the instrument used for the determination of heat of combustion in the laboratory.

Answer: Bomb calorimeter

(iv)When 0.06 mole of methane is burnt in air, the heat liberated raises the temperature of 100g of water from 70°C to 85°C.
What is the heat of combustion of methane? [ specific heat capacity = 4.2Jg-1 °C-1]

Solution:

Q = mC∆T

Q = 100 x 4.2 x (85 – 70)

= 6300Jmol-1

= 6.3KJmol-1

Question 3

Define the following terms

i. Standard heat of formation 

Answer: This is the heat absorbed or evolved when one mole of substance is formed from its elements under standard conditions.

ii. Heat of solution

Answer: This is the heat evolved or absorbed when one mole of a substance is completely dissolved in a large volume of solvent to form a solution of stated concentration.

iii. Heat of neutralization

Answer: This is the amount of heat evolved when 1mole of hydrogen ion from an acid reacts with 1mole of hydroxyl ion from an alkali to form 1mole of water, under standard conditions.

iv. Heat of combustion

Answer: This is the amount of heat evolved or liberated when 1mole of substance is burnt completely in excess oxygen under standard conditions.

Question 4

(a). Explain why the heat of neutralization between strong acid and strong alkali pair is of an approximate value of -57KJmol-1.

Answer: This is because there is a complete dissociation of thhe acid and base, while salt is formed.

(b). when 1 mole of hydrochloric acid neutralized 1 mole of potassium hydroxide, -57.5KJ of heat was given off. Calculate the heat evolved when

i. 1.5 mole of hydrochloric acid

Solution

1mole of HCl and 1mole of POH gives -57.5kJ

1.5mole of HCl

x = 1.5  x- 57KJ

= 86.25KJ

ii. 3 moles of hydrochloric acid react with excess potassium hydroxide solution.

Solution:

1mole of HCl  – 57.5KJ

3moles of HCl   – x

x = 3 x -57.5

= 172.5KJ

Question 5

[a] The heat of formation of carbon(IV)oxide is -382KJ/mole. Calculate the heat given off awhen carbon is burnt to form 8g of carbon[iv]oxide gas. [C = 12, O = 16]

Solution:

8g  →  -382KJ

CO2 = 12 + (16 x 2) = 44g

1g → \(\frac{-382}{8} \scriptsize KJ\)

44g = \(\frac{-382}{8} \times \frac{44}{1}\)

= \( \scriptsize 2101KJ \)


[b] The enthalpy change of a reaction is as follows:
CH4[g] + 2O2[g] → CO2[g] + 2H2O[g]   is -890.7KJmol-1

Given that
∆Hθ[CO2] = -193.3KJmol-1
∆Hθ[H2O] = -286.1KJmol-1

Calculate the heat of combustion of methane

Solution:

Here Oxygen is one of the reactants we ignore the enthalpy change of that since we can’t find the value of oxygen being burnt in oxygen (doesn’t make sense so we don’t have a value for the combustion of oxygen.)

∆H = ∑∆H products – ∑∆H reactants

-890.7KJ = ∑∆H products – ∑∆H reactants

= ∆Hθf  [CO2 ] + ∆Hθ 2[H2O] – ∆Hθc[CH4] = -890.7KJmol-1

-193.3 + 2(-286.1) – ∆Hθc[CH4] = -890.7KJmol-1

-193.3  – 572   – ∆Hθc[CH4] = -890.7mol-1

-765.3 – ∆Hθc[CH4] = -890.7

-∆HΘ[CH4] =  – 890.7 + 765.3KJmol-1

= -125.4KJmol-1


[c] In an experiment, 2 g of methanol was burned completely in air.

i. Write an equation for the combustion of methanol

Answer: CH3OH(l) + 1½O2(g) CO2(g) + 2H2O(l)

ii. Using your equation and the data given above, calculate the ∆cH (standard
enthalpy change of combustion) for methanol.
The enthalpy changes of formation are as follows:

Substance Standard Enthalpy Change of Formation
(∆fH) / kJ mol–1
CH3OH(l) –234
CO2(g) –394
H2O(l) –286

 

Solution:

∆cH = Σv∆fH(products) – Σv∆fH(reactants)
∆cH = (–394 + (2 x –286))kJ mol–1 – (–234 kJ mol–1 )
∆cH = –966 kJ mol–1 + 234 kJ mol–1 )
∆cH = –732 kJ mol–1

iii. Comment on whether this is an endothermic or exothermic reaction (using
your answer to (b))

Answer:  This reaction is exothermic. The negative sign indicates that heat is released during the reaction

Question 6

The enthalpy change of formation of NaCl was under investigation. 75 cm3 of 1 M
HCl was poured into a beaker then 75 cm3 NaOH 1 M was added to the HCl.

Data:

The enthalpy changes of formation are as follows:

Substance Standard Enthalpy Change of Formation / kJ mol–1
H2O(l) –286
HCl(aq) –167
NaOH(aq) –425
NaCl(aq) –411

Solution:

(a) Write an equation for the neutralisation of HCl by NaOH.

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)


(b) Using your equation and the data given above, calculate the ∆rH of this
reaction

Solution:

∆rH = Σv∆fH(products) – Σv∆fH(reactants)
∆rH = (–411 + –286) kJ mol–1 – (–425 + –167) kJ mol–1
∆rH = –697 k J mol–1 – (–592) kJ mol–1
∆rH = –697 k J mol–1 + 592 kJ mol–1
∆rH = –105 kJ mol–1

(c) Comment on the sign and magnitude of the value calculated in part (b).

Answer:

The negative sign indicates that heat is created and given out during the reaction and consequently, the reaction is exothermic.

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