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121. The equation that represents enthalpy of atomization of hydrogen is:

½H₂O(g) → H(g) +218 kJ mol⁻¹
½H₂(g) → H(g) +218 kJ mol⁻¹
H₂O(g) → 2H(g) +218 kJ mol⁻¹
½H₂(g) → H(g) –218 kJ mol⁻¹

122. Standard enthalpy of combustion of graphite at 25°C is –395.41 kJ/mol and that of diamond is –393.5 kJ/mol. The enthalpy change of graphite to diamond is:

–1.91
2.1
2.91
1.91

123. ½ H2 (g) →H(g) ΔH =218 kJ/mol in this reaction ΔH will be called

Anthalpy of atomization
Anthalpy of above de-composition
Anthalp of formation
Anthalpy of association

124. Mg + ½O₂ → MgO(g) ΔH = –692 kJ/mol at STP. Enthalpy of above reaction will be called:

ΔH⁰ₐₜ
ΔH⁰ₛ
ΔH⁰ₛₒₗ
ΔH⁰f

125. Which one of the following enthalpy changes is always exothermic?

Enthalpy of combustion
Enthalpy of solution
Enthalpy of atomization
Enthalpy of fusion

126. Which of the equations shows twice the enthalpy change of neutralization as the following equation: HCl + NaOH → NaCl + H₂O

NH₄Cl + NaOH → NaCl + H₂O + NH₃
MgCO₃ + 2HCl → MgCl₂ + CO₂ + H₂O
KOH + HCl → KCl + H₂O
H₂SO₄ + Mg(OH)₂ → MgSO₄ + 2H₂O

127. The given diagram shows the enthalpy changes during a chemical reaction. The diagram represents:

A non-spontaneous process
An isothermic process
An endothermic reaction
An exothermic reaction

128. Which enthalpy change is relevant in the following process: Na(s) → Na(g) ΔH = ?

Enthalpy of atomization
Enthalpy of vaporization
Enthalpy of fusion
Enthalpy of formation

129. The enthalpy change when one mole of water is formed by the reaction of acid with an alkali under standard conditions is known as:

Enthalpy of formation
Enthalpy of reaction
Enthalpy of combustion
Enthalpy of neutralization

130. Born–Haber cycle is used to determine the lattice energy of ionic compounds. It is the application of:

Henry's law
Le-Chatelier's principle
Hess's law
Common Ion Effect

131. All of the following steps are used to calculate the lattice energy in Born–Haber cycle except:

Atomizing the metal
Ionizing the metal
Deionize the metal
Ionize non-metal

132. For an equilibrium reaction: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) The forward reaction is exothermic, increase in temperature shifts the equilibrium position toward left because:

The concentrations of SO₂ and O₂ decrease and concentration of SO₃ increases as the temperature increases
The concentration of SO₂ and O₂ increases and concentration of SO₃ decreases as the temperature increases
The concentrations of SO₂ and O₂ increase and concentrations of SO₃ stays same as the temperature increases
The concentrations of SO₂, SO₃ and O₂ increase as the temperature increases

133. What is the ultimate fate of reversible reaction?

Completion of reaction
Complete consumption of products
Complete consumption of reactants
A state when there is no net concentration change

134. In reversible reaction, when product is removed, the equilibrium shifts towards the:

Reactant side
Product side
Both side one by one
No effect

135. Formation of NH₃ is a reversible and exothermic process. What will happen on cooling?

More reactant will form
More H₂ will form
More N₂ will form
More Product NH₃ will form

136. The value of equilibrium constant (Kc) for the reaction 2HF ⇌ H₂ + F₂ is 10⁻¹³ at 2000°C. Calculate the value of Kp for this reaction:

2 × 10⁻¹³
10⁻¹³
186 × 10⁻¹³
3.48 × 10⁻⁹

137. The catalyst used for the manufacture of H₂SO₄ by the contact process is:

SO₂
V₂O₅
Fe₂O₃
Pt/Pd

138. One can estimate the direction in which equilibrium will shift with the help of:

Le Chatelier's Principle
Law of Mass Action
Hess's Law
Present in any amount

139. The principle that states that if a stress is applied to a system at equilibrium, the system nullifies the effect of stress is:

Haber's
Le-Chatelier's
Boyle's
Charles'

140. Identify the CORRECT option required for maximum yield of ammonia by Haber's process:

High pressure, low temperature, continual removal of NH₃
Low pressure, low temperature, continual removal of NH₃
High pressure, high temperature, continual removal of NH₃
High pressure, low temperature, continual addition of NH₃