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1321. Heat of formation (ΔH°f) for CO₂ is:

–394 kJ/mol
394 kJ/mol
–294 kJ/mol
–390 kJ/mol

1322. 2H₂O → ΔH = +285.5 kJ/mol What will be the enthalpy change in the equation?

285.5 kJ/mol
Zero kJ/mol
–285.5 kJ/mol
1 kJ/mol

1323. Combustion of graphite to form CO₂ can be done by two ways. Reactions are: C + O₂ → CO₂ ΔH = –284 kJ/mol C + ½O₂ → CO ΔH = –110 kJ/mol CO + ½O₂ → CO₂ ΔH = –676 kJ/mol What will be enthalpy of formation of CO?

–110 kJ/mol
–110 kJ/mol
110 kJ/mol
676 kJ/mol

1324. Reaction of water with quick lime results in the rise of temperature of the system. Using the concept of energy change, indicate the nature of the reaction:

Third-order reaction
Non-spontaneous reaction
Endothermic reaction
Exothermic reaction

1325. 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

1326. 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

1327. 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

1328. Mg + ½O₂ → MgO(g) ΔH = –692 kJ/mol at STP. Enthalpy of above reaction will be called:

ΔH⁰ₐₜ
ΔH⁰ₛ
ΔH⁰ₛₒₗ
ΔH⁰f

1329. Which one of the following enthalpy changes is always exothermic?

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

1330. 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

1331. 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

1332. 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

1333. 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

1334. 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

1335. 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

1336. 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

1337. 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

1338. In reversible reaction, when product is removed, the equilibrium shifts towards the:

Reactant side
Product side
Both side one by one
No effect

1339. 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

1340. 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