NEETshala
Log in Get Started Free

NEET Practice Questions (MCQs) with Answers & Solutions

Practice free NEET NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

All Physics Chemistry Botany Zoology
Register free to filter questions
● ● ● ● ●

Match the following Column–I A. Isothermal process B. Adiabatic process C. Isobaric process D. Isochoric process Column–II P. $ q = \triangle U $ Q. $ w = – P \triangle V $ R. $ w = \triangle U $ S. $ w = –n RT ln (V_2/V_1) $

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

Let's match the processes:

A. Isothermal process (constant temperature) - Work done $ w = –n RT ln (V_2/V_1) $ [S]

B. Adiabatic process (no heat exchange) - Work done $ w = \triangle U $ [R]

C. Isobaric process (constant pressure) - Work done $ w = – P \triangle V $ [Q]

D. Isochoric process (constant volume) - Heat added equals change in internal energy $ q = \triangle U $ [P]

● ● ● ● ●

Reasoning question choose the correct statement. Statement–1 : The enthalpy of formation of $H_2O(l)$ is greater than of $H_2O (g) $ . Statement–2: Enthalpy change is negative for the condensation reaction $H_2O (g) \rightarrow H_2O(l) $

You've reached today's free limit of 20 questions. Log in to keep practising for free.
● ● ● ● ●

Reasoning question choose the correct statement. Statement–1 : Heat of neutralisation of perchloric acid, $HClO_3$, with NaOH is same as that of HCl with NaOH. Statement–2: Both HCl and $HClO_4$ are strong acids.

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

Statement –1 is true because the heat of neutralization of perchloric acid, $HClO_4$, with NaOH is the same as that of HCl with NaOH. This is because both $HCl$ and $HClO_4$ are strong acids and completely dissociate in water. Statement –2 is also true because both $HCl$ and $HClO_4$ are strong acids. Therefore, Statement –2 correctly explains Statement –1.

● ● ● ● ●

Reasoning question choose the correct statement. Statement–1 : When a gas at high pressure expands against vacuum, the work done is maximum. Statement–2: Work done in expansion depends upon the pressure inside the gas and increase in volume.

You've reached today's free limit of 20 questions. Log in to keep practising for free.
● ● ● ● ●

Reasoning question choose the correct statement. Statement I : The chemical reaction, $3H_2(g) + N_2(g) \rightarrow 2NH_3 $ shows decrease in entropy. Statement II: The process passes into equilibrium state when $ \triangle GT,P $ becomes zero.

You've reached today's free limit of 20 questions. Log in to keep practising for free.
● ● ● ● ●

Calculate the work performed when 2 moles of hydrogen expand isothermally and reversibly at $25 ^\circ C $ form 15 to 50 litres.

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

W = – 2.303 n RT log $ { V_2 \over V_1 } = - 2.303 \times 2 \times 2 \times 298 \times log { 50 \over 15 } = -1436 calories $

● ● ● ● ●

The $ \triangle H_f0 for CO_2(g), CO(g) and H_2O (g)$ are –393.5, –110.5 and –241.8 $kJ mol^{–1} $ respectively. The standard enthalpy change (in kJ) for the reaction $CO_2(g) + H_2(g) \rightarrow CO(g) + H_2O (g) $ is –

You've reached today's free limit of 20 questions. Log in to keep practising for free.
● ● ● ● ●

For reaction carried out in automobiles, what is the value of $ \triangle H , \triangle S and \triangle G ? $ $ 2C_8H_{18} (g) + 25O_2(g) 16CO_2(g)+ 18H_2O(g) $

You've reached today's free limit of 20 questions. Log in to keep practising for free.
● ● ● ● ●

At $25 ^\circ $ temperature equilibrium constant Kp for given reaction is $ = 1.8 10 ^{–7} $ Then what is the value of $ \triangle G ^\circ ? PCl_5 \rightleftharpoons PCl_3 + Cl_2 $

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

$ \triangle G ^\circ = – RT ln Kp $ = – 2.303RT log Kp $ = – 2.303 298 log 1.8 10^ {–7} $ = – 2.303 1.987 298 (– 6.7447) = 9197.5 Cal

● ● ● ● ●

The value of $ \triangle H_f ^\circ of U_3O_8 is –853.5 KJ mol^{–1} $ . Also $ \triangle H ^\circ $ for the reaction $3UO_2 + O_2 \rightarrow U_3O_8, $ is –76.00 KJ. The value of $ \triangle H_f ^\circ of UO_2 $ is approx –

You've reached today's free limit of 20 questions. Log in to keep practising for free.
Explanation

To find the enthalpy of formation ($ riangle H_f^ ext{°}$) of $UO_2$, we can use the given data and the enthalpy change of the reaction. The given enthalpy changes are:

$ riangle H_f^ ext{°}(U_3O_8) = -853.5$ kJ/mol and $ riangle H^ ext{°}$ for the reaction $3 UO_2 + O_2 ightarrow U_3O_8$ is $-76.00$ kJ.

Using Hess's law, we can write the enthalpy change for the reaction as:

$ riangle H^ ext{°} = riangle H_f^ ext{°}(U_3O_8) - 3 riangle H_f^ ext{°}(UO_2) - riangle H_f^ ext{°}(O_2)$

Since the enthalpy of formation of $O_2$ in its standard state is zero, we have:

$-76.00 = -853.5 - 3 riangle H_f^ ext{°}(UO_2)$

Rearranging to solve for $ riangle H_f^ ext{°}(UO_2)$:

$3 riangle H_f^ ext{°}(UO_2) = -853.5 + 76.00$

$3 riangle H_f^ ext{°}(UO_2) = -777.5$

$ riangle H_f^ ext{°}(UO_2) = -777.5 / 3 = -259.17$ kJ/mol

Thus, the value of $ riangle H_f^ ext{°}$ of $UO_2$ is approximately $-259.17$ kJ/mol.

Ready to ace NEET?

Free access · No credit card required

Browse All Questions Create Free Account

Frequently Asked Questions

Yes. You can attempt every NEET question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.