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Chemical Thermodynamics Mock Tests

45 questions available

Chemical Thermodynamics Mock Test 1

Questions: 30

Chemical Thermodynamics Mock Test 2

Questions: 15

Sample Questions

NEET Chemistry
Given: C(s) + O₂(g) → CO₂(g), ΔH = -393.5 kJ mol⁻¹; CO(g) + ½O₂(g) → CO₂(g), ΔH = -283.0 kJ mol⁻¹. The enthalpy of formation of CO(g) is:
A -110.5 kJ mol⁻¹
B -676.5 kJ mol⁻¹
C -1105 kJ mol⁻¹
D -69.5 kJ mol⁻¹
NEET Chemistry
Given: C(s) + O₂(g) → CO₂(g), ΔH = -393.5 kJ/mol 2CO(g) + O₂(g) → 2CO₂(g), ΔH = -566.0 kJ/mol What is the enthalpy of formation of CO(g) from C(s) and O₂(g)?
A -110.5 kJ/mol
B -283.0 kJ/mol
C -676.5 kJ/mol
D -196.75 kJ/mol
JEE Main Chemistry
The enthalpy of combustion of carbon to CO₂ is -393.5 kJ mol⁻¹. The mass of carbon required to produce 1570 kJ of heat at STP is:
A 12 g
B 24 g
C 36 g
D 48 g
NEET Chemistry
Which of the following statements is correct about the second law of thermodynamics?
A Energy can neither be created nor destroyed
B The entropy of the universe tends to a maximum
C Entropy of a perfect crystal at 0 K is zero
D Heat cannot flow from cold to hot without external work
JEE Main Chemistry
The enthalpy of neutralization of a strong acid by a strong base is −57.1 kJ/mol. The enthalpy of neutralization of HCN by NaOH will be:
A Less than −57.1 kJ/mol (more negative)
B Exactly −57.1 kJ/mol
C Greater than −57.1 kJ/mol (less negative)
D Zero
JEE Main Chemistry
The heat of reaction at constant pressure (Delta H) is related to Delta E by:
A Delta H = Delta E + Delta nRT
B Delta H = Delta E + RT
C Delta H = Delta E - Delta nRT
D Delta H = Delta nRT
JEE Main Chemistry
The entropy change (ΔS) for the vaporization of a liquid is 108 J K⁻¹ mol⁻¹. If its enthalpy of vaporization is 30.8 kJ mol⁻¹, the boiling point of the liquid is:
A 285.2 K
B 285.2°C
C 308 K
D 273 K
JEE Main Chemistry
When 1 mole of an ideal gas at 25°C is expanded reversibly from 8 dm³ to 20 dm³, the value of q (heat absorbed) is:
A 8.89 kJ
B 4.45 kJ
C 17.78 kJ
D 2.23 kJ

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