JEE Main Thermodynamics Practice Questions With Solutions

Given below are two statements: One is labelled as Assertion (A) and the other is labelled as Reason (R)

Assertion (A) : Enthalpy of neutralisation of strong monobasic acid with strong monoacidic base is always $-57\mathrm{kJ}{\mathrm{mol}}^{-1}$- 57 \textrm{ } kJ \left(\textrm{ } mol\right)^{- 1}

Reason (R) : Enthalpy of neutralisation is the amount of heat liberated when one mole of ${\mathrm{H}}^{+}$H^{+} ions furnished by acid combine with one mole of ${}^{-}\mathrm{OH}$^{-} OH ions furnished by base to form one mole of water.

In the light of the above statements, choose the correct answer from the options given below.

Which of the following is not correct?

What happens when methane undergoes combustion in systems A and B respectively?

Given

$\mathrm{\Delta }{\mathrm{H}}^0$\Delta H^{0} for the reaction

$\mathrm{C}(\mathrm{graphite})+\frac{1}{2}{\mathrm{O}}_2(\mathrm{g})\to \mathrm{CO}(\mathrm{g})$C \left(\right. graphite \left.\right) + \frac{1}{2} O_{2} \left(\right. g \left.\right) \rightarrow CO \left(\right. g \left.\right) is :

Which of the following relations are correct?

(A) $\mathrm{\Delta }\mathrm{U}=\mathrm{q}+\mathrm{p}\mathrm{\Delta }\mathrm{V}$\Delta U = q + p \Delta V

(B) $\mathrm{\Delta }\mathrm{G}=\mathrm{\Delta }\mathrm{H}-\mathrm{T}\mathrm{\Delta }\mathrm{S}$\Delta G = \Delta H - T \Delta S

(C) $\mathrm{\Delta }\mathrm{S}=\frac{q_{rev}}{T}$\Delta S = \frac{q_{r e v}}{T}

(D) $\mathrm{\Delta }\mathrm{H}=\mathrm{\Delta }\mathrm{U}-\mathrm{\Delta }\mathrm{nRT}$\Delta H = \Delta U - \Delta nRT

Choose the most appropriate answer from the options given below :

Given below are two statements: One is labelled as Assertion $\mathbf{A}$\mathbf{A} and the other is labelled as Reason $\mathbf{R}$\mathbf{R}

Assertion A : The reduction of a metal oxide is easier if the metal formed is in liquid state than solid state.

Reason $\mathbf{R}$\mathbf{R} : The value of $\mathrm{\Delta }{G}^{\mathrm{\Theta }}$\Delta G^{\Theta} becomes more on negative side as entropy is higher in liquid state than solid state.

In the light of the above statements, choose the most appropriate answer from the options given below

Which of the following relation is not correct?

$\mathrm{\Delta }$\DeltaG${}^{\circ }$^{\circ} vs T plot for the formation of MgO, involving reaction

2Mg + O₂$\to$\rightarrow 2MgO, will look like :

Match List-I with List-II.

Choose the correct answer from the options given below :