Theromodynamics and Thermochemistry Test

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Theromodynamics and Thermochemistry Test - 3

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Weekly Quiz Competition

Question 1
1 / -0

The mutual heat of neutralisation of 40 g NaOH and 60 g CH₃COOH will be–

A
57.1 kJ

B
less than 57.1 kJ

C
more than 57.1 kJ

D
13.7 kJ

Solution

Heat of neutralisation of strong acid & base = 57.1 kJ
Since NaOH and CH₃COOH react together so heat of neutralisation will be less than 57.1 kJ as NaOH is a strong base and CH₃COOH is a weak base as some of the heat is utilised to ionise acetic acid.
Question 2
1 / -0

Which one is not a state function -

A
Internal energy (E)

B
Volume

C
Heat (q)

D
Enthalpy

Solution

State functions are those function whose values depend only upon the initial and final states of the system not on the path.
Question 3
1 / -0

One mole of an ideal gas at 300 K is expanded isothermally from an initial volume of 1 litre to 10 liters. The ΔE for this process is
(R = 2 cal mol^–1 K^–1)–

A
163.7 cal

B
zero

C
138.1 cal

D
9 lit atm

Solution

For isothermal process ΔE = 0 because ΔT = 0
Question 4
1 / -0

From the standard enthalpies of formation values for some compounds (in KJ mole^–1), predict which one of them will be most stable?

A
Propane (g) (ΔH_f º = – 103.8)

B
n-Butane (g) (ΔH_f º = – 124.7)

C
n-Butane (g) (ΔH_f º = – 167.2)

D
n-Octane (g) (ΔH_f º = – 208.4)

Solution

Greater the standard enthalpy of formation compound will be more stable
Question 5
1 / -0

A gas expands isothermally and reversibly. The work done by the gas is -

A
Zero

B
Minimum

C
Maximum

D
Cannot be determined

Solution

W_rev > W_irev.
Work done by the gas is maximum

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Theromodynamics and Thermochemistry Test - 3

Result

Theromodynamics and Thermochemistry Test - 3

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Question 1

57.1 kJ

less than 57.1 kJ

more than 57.1 kJ

13.7 kJ

Solution

Question 2

Internal energy (E)

Volume

Heat (q)

Enthalpy

Solution

Question 3

163.7 cal

zero

138.1 cal

9 lit atm

Solution

Question 4

Propane (g) (ΔHf º = – 103.8)

n-Butane (g) (ΔHf º = – 124.7)

n-Butane (g) (ΔHf º = – 167.2)

n-Octane (g) (ΔHf º = – 208.4)

Solution

Question 5

Zero

Minimum

Maximum

Cannot be determined

Solution

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Propane (g) (ΔH_f º = – 103.8)

n-Butane (g) (ΔH_f º = – 124.7)

n-Butane (g) (ΔH_f º = – 167.2)

n-Octane (g) (ΔH_f º = – 208.4)