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QUESTION: 1

**Direction (Q. Nos. 1-7) This section contains 7 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.**

**Given,**

The enthalpies of elements in their standard states are taken as zero. The enthalpy of formation of a compound

Solution:

Depend on whether the formation of compounds is exothermic or endothermic.For exothermic reaction, enthalpy of formation would be negative and for endothermic,it is positive.

QUESTION: 2

Given,

Solution:

Applying i-ii/2-3/2iii

We get ∆_{f}H of NCl_{3} in terms of ∆H1, ∆H_{2} and ∆H_{3}

QUESTION: 3

Diborane is a potential rocket fuel which undergoes combustion according to the reaction,

From the following data, enthalpy change for the combustion of diborane is

Solution:

On applying equation 1 + 3 times equation 2 minus equation 4 plus equation 3, we will get enthalpy change for the combustion of diborane.

QUESTION: 4

Following diagram represents Born-Haber cycle to determine lattice energy of NaCI(s). It is based on Hess’s law of constant heat summation. Δ_{lattice}H° of

Solution:

The correct answer is Option A.

-411.2 = 108.4 + 495.6 + 121.0 - 348.6 + LE

=>LE= -411.2 - 108.4 - 495.6 - 121.0 +348.6

=> LE = -787.6 ~ -788 kJ mol^{-1}

QUESTION: 5

Thus,

Solution:

Put the things we need as reactant on LHS the product on RHS. To do that, we will apply equation S+I+D-2E-U.

QUESTION: 6

Given, HCI (g) → H (g) + Cl (g) at 298 K (say temperature T K),

For

**Q. Thus, for the given reaction (at 0 K) is**

Solution:

Ho= Ho of product - Ho of reactant

so ho =(6.197 + 6.272 ) - 8.644

= 3.825

Ho298-Ho = 431.961 - 3.825=428.136532873

QUESTION: 7

The measured enthalpy change for burning of ketene (g) (CH_{2}CO) is - 981.1 kJ mo^{-1} at 298 K ,CH_{2}CO (g)+ 2O_{2}(g) → 2CO_{2} (g) + H_{2}O (g) and that of CH_{4} (g) is - 802.3 kJ mol^{-1} at 298 K

Thus, enthalpy change at 298 K for the following thermochemical reaction is

Solution:

CH_{2}CO (g)+ 2O_{2}(g) → 2CO_{2} (g) + H_{2}O (g) -(i)

CH4 (g)+ 2O_{2}(g) → CO_{2} (g) + 2H_{2}O (g) -(ii)

On 2(ii)-(i)

2CH_{4} (g)+ 2O_{2}(g) → CH_{2}CO (g) + 3H_{2}O (g) -(iii)

∆H_{(iii)} = 2∆H_{(ii)} - ∆H_{(i)} = [2 (-802.3) -(-981.1)] -623.5 kJ mol^{-1}

QUESTION: 8

Direction (Q. Nos. 8) This sectionis based on statement I and Statement II. Select the correct answer from the code given below.

**Q. In the following thermochemical reactions,**

Statement I : Enthalpy of combustion of CO(g) is - 26 kcal mol^{-1}.

Statement II : CO_{2}(g) is an exothermic compound.

Solution:

According to me, the question is presented in an improper way. For statement I to be correct, value of enthalpy of combustion of CO be -68 kcal mol^{-1}. This can be sensed by reversing reaction II adding it to reaction I. By this, we will get the value of enthalpy of combustion of CO be -68 kcal mol^{-1}.

However, statement II is wrong. There is no such exothermic or endothermic compounds. Reactions are exothermic and not the products.

QUESTION: 9

**Direction (Q. Nos. 9 and 10) This section contains a paragraph, wach describing theory, experiments, data etc. three Questions related to paragraph have been given.Each question have only one correct answer among the four given ptions (a),(b),(c),(d).**

**Consider the following thermochemical equations**

**Q. Δ _{r}H° of the following thermochemical reaction is**

**CIO _{2}**

Solution:

The question can be simply done by understanding that this reaction is endothermic so ∆H is negative.

**Hence A is correct.**

QUESTION: 10

**Consider the following thermochemical equations**

**Q. Δ _{r}H° of the following thermochemical reaction is**

Solution:

QUESTION: 11

**Direction (Q. Nos. 11) Choice the correct combination of elements and column I and coloumn II are given as option (a), (b), (c) and (d), out of which ONE option is correct.**

**Q. For liquid water at **

Compare the parameters ΔH_{1}, ΔH_{2} ... in Column I with the corresponding values in kJ in Column II.

Solution:

∆H^{1}=∆rH298=-285.83kJmol-¹

∆H²=Cp(liq)∆T=75.29×75 =5.64kJmol-¹

∆H³=∆Hvap =40.88kJmol-¹

∆H⁴=Cp(gas)∆T=35.57×(-75) =-2.5183kJmol-¹

*Answer can only contain numeric values

QUESTION: 12

**Direction (Q. Nos. 12-15) This section contains 4 questions. when worked out will result in an integer from 0 to 9 (both inclusive)**.

**Q. Based on (BE) values, Δ _{r}H° of the following reaction 298 K is**

Solution:

Let the BE value of B---Br bond be x and B---Cl bond be y

BE of BBr3 is 3x, BCl3=3y , BBr2Cl=2x+y ,BCl2Br=2y+x

then ∆Hr=Hr(products)-Hr(reactants)=(2x+y+2y+x) - (3x+3y)=0

*Answer can only contain numeric values

QUESTION: 13

Heat of combustion of CH_{2}CO(g) is - 981.1 kJ mol^{-1} and that of CH_{4} (g) is - 802.3 kJat 298 K. If 1247.0 kJ of heat is released in the following change

**Q. How many moles of CH _{4}(g)are used?**

Solution:

[2CH_{4 }+ 2O_{2} → CH_{2}O + H_{2}O]n

∆H_{C} = n[-981.1-(-8023×2)]

1247 = n[-981.1+1604.6]

N[623.5] = 1247

N = 4

So, no of moles of CH_{4} = n×2 = 2×2 = 4

*Answer can only contain numeric values

QUESTION: 14

Consider the following reactions

**Q. What is the resonance energy (in kcal) of A?**

Solution:

Resonance energy = (Observed ∆H) - (Theoretical ∆H)

= (-98.0) - 2(-45.0)

= -8 kcal mol^{-1}

So we have Resnance energy = 8

*Answer can only contain numeric values

QUESTION: 15

Given, enthalpy of combustion of carbon (s) = -94 kcal mol^{-1}

H_{2}(g) = - 68 kcal mol^{-1} and C_{X}H_{4} = - 318 kcal mol^{-1} and Δ_{f}H° (C_{x}H_{4}) = - 100 kcal mol^{-1}

**Q. What is the value of x?**

Solution:

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