Thermodynamics Test

Result

Thermodynamics Test - 5

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Question 1
1 / -0

A reaction, A + B → C + D + q is found to have a positive entropy change. The reaction will be

A
spontaneous at all temperature.

B
spontaneous only at high temperature.

C
spontaneous only at low temperature.

D
nonspontaneous at all temperature.

Solution

ΔG = ΔH -T ΔS

ΔS is positive and ΔH is negative as heat is liberated in the reaction.

so ΔG is negative hence reaction will be spontaneous at all temperature.
Question 2
1 / -0

In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?

A
ΔU = 327 J

B
ΔU = 367 J

C
ΔU = 307 J

D
ΔU = 347 J

Solution

ΔU = q + w (1st law of thermodynamics)

q = +701J (heat is absorbed by system)

w = -394 J (work is done by system)

ΔU = 701- 394

= 307J
Question 3
1 / -0

Enthalpy of combustion of carbon to CO₂ is 393.5 kJ mol^-1. Calculate the heat released upon formation of 35.2 g of CO₂from carbon and dioxygen gas.

A
-315 kJ

B
-398 kJ

C
-375 kJ

D
-275 kJ

Solution

When 1 mole of CO₂is produced energy released is –393.5 kJ mol^-¹. Moles of CO₂ given = 35.2/44 = 0.8 moles; So energy released =0.8×393.5 kJ/mol = 315 kJ/mol