back to directory
NEET CHEMISTRYThermodynamicsMedium

Question

ΔE∘\Delta E^{\circ} of combustion of isobutylene is −X kJ mol−1-X \text{ kJ mol}^{-1}. The value of ΔH∘\Delta H^{\circ} is [DCE 2004]:

A

=ΔE∘= \Delta E^{\circ}

B

>ΔE∘> \Delta E^{\circ}

C

=0= 0

D

<ΔE∘< \Delta E^{\circ}

Step-by-Step Solution

To determine the relationship between ΔH∘\Delta H^{\circ} and ΔE∘\Delta E^{\circ} (internal energy change), we first write the balanced thermochemical equation for the combustion of isobutylene (2-methylpropene, C4H8C_{4}H_{8}). Based on general combustion principles provided in the sources, the reaction is:

C4H8(g)+6O2(g)→4CO2(g)+4H2O(l)C_{4}H_{8}(g) + 6O_{2}(g) \rightarrow 4CO_{2}(g) + 4H_{2}O(l)

According to the sources, the relationship between enthalpy and internal energy is ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta n_{g}RT . Here, Δng\Delta n_{g} is the change in the number of moles of gaseous products and reactants .

  1. Gaseous products: 4 moles of CO2(g)CO_{2}(g). (Note: H2OH_{2}O is produced as a liquid under standard conditions).
  2. Gaseous reactants: 1 mole of C4H8(g)C_{4}H_{8}(g) and 6 moles of O2(g)O_{2}(g), totaling 7 moles.
  3. Calculate Δng\Delta n_{g}: Δng=4−7=−3\Delta n_{g} = 4 - 7 = -3 .

Substituting this into the formula yields: ΔH∘=ΔE∘+(−3)RT\Delta H^{\circ} = \Delta E^{\circ} + (-3)RT, which simplifies to ΔH∘=ΔE∘−3RT\Delta H^{\circ} = \Delta E^{\circ} - 3RT.

Since RR and TT are positive values, subtracting 3RT3RT from ΔE∘\Delta E^{\circ} results in a value that is smaller. Therefore, ΔH∘<ΔE∘\Delta H^{\circ} < \Delta E^{\circ}, which matches Option D.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Thermodynamics. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYThermodynamicscombustionisobutylene

More Thermodynamics Questions

View all

For the gas phase reaction, $\text{PCl}_5\text{(g)} \rightleftharpoons \text{PCl}_3\text{(g)} + \text{Cl}_2\text{(g)}$, which of the following conditions is correct?

A.$\Delta H = 0$ and $\Delta S < 0$
B.$\Delta H > 0$ and $\Delta S > 0$
C.$\Delta H < 0$ and $\Delta S < 0$
D.$\Delta H > 0$ and $\Delta S < 0$
EasySolve

One mole of an ideal gas at $300\text{ K}$ is expanded isothermally from an initial volume of $1\text{ litre}$ to $10\text{ litres}$. The $\Delta E$ for this process is: ($R = 2\text{ cal mol}^{-1}\text{K}^{-1}$)

A.$1381.1\text{ cal}$
B.Zero
C.$163.7\text{ cal}$
D.$9\text{ lit. atm}$
EasySolve

What is the amount of work done by an ideal gas, if the gas expands isothermally from $10^{-3} \text{ m}^3$ to $10^{-2} \text{ m}^3$ at $300 \text{ K}$ against a constant pressure of $10^5 \text{ N m}^{-2}$?

A.$+270 \text{ kJ}$
B.$-900 \text{ J}$
C.$+900 \text{ kJ}$
D.$-900 \text{ kJ}$
MediumSolve

Match List-I (Equations/Conditions) with List-II (Type of processes) and select the correct option. **List-I (Equations)** (a) $K_p > Q$ (b) $T < \frac{\Delta H}{\Delta S}$ (for $\Delta H > 0$) (c) $K_p = Q$ (d) $T > \frac{\Delta H}{\Delta S}$ (for $\Delta H > 0$) **List-II (Type of processes)** (i) Non-spontaneous (ii) Equilibrium (iii) Spontaneous and endothermic (iv) Spontaneous

A.(a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)
B.(a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
C.(a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
D.(a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
MediumSolve

The enthalpy of neutralization of which of the following acids and bases is nearly –13.6 Kcal [Roorkee 1999]

A.HCN and NaOH
B.HCl and KOH
C.HCl and HCN
D.HCl and NH4OH
EasySolve

The heat of combustion of carbon to $\text{CO}_2$ is $-393.5 \text{ kJ/mol}$. The heat released upon the formation of $35.2 \text{ g}$ of $\text{CO}_2$ from carbon and oxygen gas is:

A.$-315 \text{ kJ}$
B.$+315 \text{ kJ}$
C.$-630 \text{ kJ}$
D.$+630 \text{ kJ}$
MediumSolve

A system is changed from state A to state B by one path and from B to A by another path. If $E_{1}$ and $E_{2}$ are the corresponding changes in internal energy, then [Pb. PMT 2001]:

A.$E_{1} + E_{2} = -ve$
B.$E_{1} + E_{2} = +ve$
C.$E_{1} + E_{2} = 0$
D.None of the above
EasySolve

The work done to contract a gas in a cylinder is 462 joules. 128 joule energy is evolved in the process. What will be the internal energy change in the process? [MP PMT 2003]

A.+ 590 joules
B.– 334 joules
C.+ 334 joules
D.– 590 joules
EasySolve

Why 32,000+ students choose TopperSquare

Everything you need to crack NEET

15,000+ Chapter MCQs

Physics, Chemistry, Biology — chapter-wise, topic-wise practice

AI Performance Analytics

Know your weak topics. Get a personalised study plan.

Full NEET Mock Tests

180 questions, timed, with detailed score analysis

PYQ Sets 2010–2024

15 years of past papers with solved explanations

Start Practising Free

No credit card · 30-second signup · Free forever plan included

This neet chemistry practice question is part of the TopperSquare free question bank. TopperSquare offers 15,000+ chapter-wise NEET MCQs across Physics, Chemistry, and Biology with detailed step-by-step explanations, full mock tests, NEET PYQs (2010–2024), and an AI-powered performance analytics dashboard. browse all neet practice questions → · practice chemistry sets →

Explore more subjects:Physics MCQs•Chemistry MCQs•Biology MCQs•Botany MCQs•Zoology MCQs•NEET Mock Tests•NEET PYQs