Thermodynamics Test

Result

Thermodynamics Test - 23

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Direction (Q. Nos. 1-11) This section contains 11 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.
Q. Calorific value of H₂ is - 143 kJ g^-1.Thus, Δ_fH° of H₂O is

A
-143 kJ mol^-1

B
-286 kJ mol^-1

C
+143 kJ mol^-1

D
+286 kJ mol^-1

Solution

Calorific value is given for 1 gram and enthalpy of formation is defined for one mole of substance formed so for one mole of water formed we need two gram of hydrogen so the enthalpy of formation of H2O will be 2×Calorific value. Or 2×(-143) = -286 kJ/ mol
Question 2
1 / -0

In the complete combustion of C₂H₆, 54 g of H₂O is formed and 370 kcal of heat is evolved. Thus, Δ_CH° of C₂H₆ is

A
370 kcal mol^-1

B
6.61 kcal mol^-1

C
-6.61 kcal mol^-1

D
-370 kcal mol^-1

Solution

54 gram of water is formed or three moles of water is formed.
C₂H₆ + (7/2)O2 → 2CO₂ + 3H₂O
So it indicates that only one mole of C₂H₆ is used in the combustion so the heat released is itself the value of heat of combustion of Ethane.
Question 3
1 / -0

The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are - 890.3 kJ mol^-1, - 393.5 kJ mol^-1 and - 285.8 kJ mol^-1 respectively. Enthalpy of formation of CH₄(g) will be

A
-74.8 kJ mol^-1

B
-52.57 kJ mol^-1

C
+74.8 kJ mol^-1

D
+52.27 kJ mol^-1

Solution
Question 4
1 / -0

The reaction of cyanamide, NH₂CN (s) was carried out in a bomb calorimeter, and ΔU was found to be - 742.7 kJ mol^-1 at 298 K. Thus, Δ_rH° at 298 K for

A
-741.5 kJ

B
+741.5 kJ

C
-743.95 kJ

D
+738.98kJ

Solution

We know, from first law of thermodynamics,
enthalpy change{∆H} = ∆U{internal energy} + ∆ngRT
Where,
∆ng = no of mole of gaseous products - no of mole of gaseous reactant
So the reaction follows :
NH₂CN(s) +(3/2)O₂(g) → N₂(g)+CO₂(g)
∆ng = np - nr
= (1 + 1) - 3/2 = 1/2 mol
= 0.5 mol
Now, ∆H = ∆U + ∆n_gRT
= -742.7 KJ +(0.5mol × 8.314 J/mol/K × 298K)
= -742 KJ + 1238.786 J
= -742 KJ + 1.238786 KJ
= -741.46 KJ
Hence, enthalpy change for the reaction is 741.46 KJ for combustion of 1 mole of NH₂CN.
Question 5
1 / -0

Based on the following thermochemical reactions at 298 K and 1 bar
Q. Enthalpy of vaporisation of H₂O (l) is

A
90 kJ mol^-1

B
45 kJ mol^-1

C
-90 kJ mol^-1

D
-45 kJ mol^-1

Solution

Enthalpy of vaporization is defined for a substance when one mole of the substance is converted into the gaseous state. To obtain the enthalpy of vaporization of H₂O liquid we will add reaction 1 and reaction 2 and then divide the enthalpy of x to since we need to calculate the enthalpy of vaporization for single mole of H₂O
Question 6
1 / -0

For the following combustion reaction of benzoic acid at 298 K
Q. Thus enthalpy of combustion Δ_CH° at 298 K is

A
3224 kJmol^-1

B
-3224 kJmol^-1

C
3225 kJmol^-1

D
-3225 kJmol^-1
Question 7
1 / -0

The standard enthalpies of formation of CO₂(g), H₂O (l) and glucose (s) at 25°C are - 400 kJ mol^-1, - 300 kJ mol^-1 and - 1300 kJ mol^-1 respectively. The standard enthalpy of combustion per gram of glucose at 25°C is
[JEE Advanced 2013]

A
2900 kJ

B
-2900 kJ

C
-16.11 kJ

D
+16.11 kJ

Solution
Question 8
1 / -0

For the complete combustion of ethanol,
the amount of heat produced as measured in bomb calorimeter is 1364.47 kJ mol^-1 at 25°C. Assuming ideality, the enthalpy of combustion, Δ_CH° for the reaction Will be (R = 8.314 JK^-1 mol^-1)
[JEE Main 2014]

A
-1366.95 kJ mol^-1

B
-1361.95 kJ mol^-1

C
-1460.50 kJ mol^-1

D
-1350.50 kJ mol^-1

Solution

∆H = ∆E + ∆n_gRT
∆E = -1364.47 kJ mol^-1and T = 298 K
∆n_g = 2-3 = -1
∆H = -1364.47 -1(8.314)289/1000 = -1366.95 kJ mol^-1
Question 9
1 / -0

Given,
Q. Thus, Δ_fH° of (C₆H₆) (l) is

A
-3267.0 kJ mol^-1

B
-48.51 kJ mol^-1

C
+3267.0 kJ mol^-1

D
+48.51 kJ mol^-1

Solution

C₆H₆(l) + 15/2O₂(g) → 6CO₂(g) + 3H₂O(l)
∆H_r = (∆H_c)_C6H6 ⇒-3261 = ∑H_product - ∑H_reactant = 6(-393.5) -3(-283.83) - ∆_fH_C6H6
∆H_f = 48 kJ mol^-1
Question 10
1 / -0

Δ_fU° of formation of CH₄(g) at a certain temperature is - 393 kJ mol^-1. The value of Δ_fH° is

A
zero

B

C

D

Solution

The balanced equation for combustion of methane is:
CH₄(g)+2O₂(g)→CO₂(g)+2H₂O(l)Here, Δn_g=1−3=−2
ΔH₀=ΔU₀+Δn_gRT
ΔH⁰=−393−2RT
∴ΔH⁰<ΔU₀
Question 11
1 / -0

Heat of combustion of H₂(g) is - 58 kcal mol^-1 at 298 K and constant pressure.
Q. Temperature of a hydrogen-oxygen flame if H₂(g) is used as fuel is

A
8638 K

B
1298 K

C
8583 K

D
1722 K