In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is $CH_3OH(l) + 3/2O2 (g) \rightarrow CO_2(g) + 2H_2O(l) $ . At 298 K, standard Gibb’s energies of formation for $ CH_3OH(l), H_2O(l) and CO_2 (g) are –166.2,–237.2 and –394.4 kJ mol^{–1} $ respectively. If standard enthalpy of combustion of methanol is $ –726kJ mol ^ {–1} $ , efficiency of the fuel cell will be :
Chemistry MCQs for NEET — Practice Questions with Answers
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The standard enthalpy of formation of $ NH_3 is – 46.0 kJ mol^{–1} $ . If the enthalpy of formation of $H_2$ from its atoms is $ –436 kJ mol^{–1} $ and that of $ N_2 is –712 kJ mol^{–1} $ , the average bond enthalpy of N – H bond in $NH_3$ is
The average bond enthalpy of N-H bonds in NH3 can be calculated using the formula: ΔHf(NH3) = Σ(Bond enthalpies of reactants) - Σ(Bond enthalpies of products) Given: ΔHf(NH3) = -46 kJ/mol ΔHf(H2 from atoms) = -436 kJ/mol ΔHf(N2 from atoms) = -712 kJ/mol Let x be the bond enthalpy of N-H. The formation of NH3 involves breaking 1/2 N2 and 3/2 H2: ΔH = 1/2(-712) + 3/2(-436) - 3x -46 = -356 - 654 - 3x 3x = -1000 + 46 3x = -1000 + 46 x = 352 kJ/mol.
Choose the correct option about the following sentnences [T= True , F =False]
(i) Ice in contact with water constitutes a homogeneous system.
(ii) The process is known as isochoric in which the pressure remains constant throughout the change, i.e., dP = 0.
(iii) A spontaneous process is reversible in nature.
(iv) In an isolated system, one form of energy cannot be converted into another, i.e., internal energy remains constant.
Choose the correct option about the following sentnences [T= True , F =False] (i)Molar heat capacity at constant pressure = Molar heat capacity at constant volume + POV. (ii)A spontaneous process is accompanied by a decrease in entropy. (iii) $ \triangle Hsub = \triangle Hfusion + \triangle Hvap $ . (iv)The standard heat of formation represents the formation of the compound from its elements at $ 25 ^\circ C $ and one atmospheric pressure. (v)Whenever an acid is neutralised by a base, the net reaction is
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) $
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) $
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.
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.
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.
Calculate the work performed when 2 moles of hydrogen expand isothermally and reversibly at $25 ^\circ C $ form 15 to 50 litres.
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 $
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