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800 most-asked Class 11 & 12 (+1 / +2) questions across Physics, Chemistry, Maths and Biology — each with a model answer and the exact marking-scheme points examiners reward. Revise smart, walk in calm.
ChemistryClass 113 marksmedium
States of Matter
A gas occupies 300 mL at 27 deg C. What volume will it occupy at 127 deg C if the pressure is kept constant?
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Convert temperatures to Kelvin: T1 = 27 + 273 = 300 K, T2 = 127 + 273 = 400 K. By Charles's law V1/T1 = V2/T2, so V2 = V1 x T2/T1 = 300 x 400/300 = 400 mL.
V1/T1 = V2/T2
Marking-scheme points
- ✓Convert to Kelvin: 300 K and 400 K
- ✓V1/T1 = V2/T2 at constant pressure
- ✓V2 = 300 x 400/300 = 400 mL
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States of Matter
Write the ideal gas equation and give the value of the gas constant R in two units.
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The ideal gas equation is PV = nRT, where P = pressure, V = volume, n = number of moles, T = absolute temperature and R = universal gas constant. R = 0.0821 L atm K^-1 mol^-1 = 8.314 J K^-1 mol^-1.
PV = nRT
Marking-scheme points
- ✓PV = nRT
- ✓R = 0.0821 L atm K^-1 mol^-1
- ✓R = 8.314 J K^-1 mol^-1
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States of Matter
Calculate the volume occupied by 2 moles of an ideal gas at 300 K and a pressure of 2 atm. (R = 0.0821 L atm K^-1 mol^-1)
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Using PV = nRT, V = nRT/P = (2 x 0.0821 x 300) / 2 = 49.26/2 = 24.63 L.
V = nRT / P
Marking-scheme points
- ✓Use V = nRT/P
- ✓Substitute n=2, R=0.0821, T=300, P=2
- ✓V = 24.63 L
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States of Matter
State the main postulates of the kinetic molecular theory of gases.
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1) A gas consists of a large number of tiny particles (molecules) whose actual volume is negligible compared with the volume of the container. 2) There are no forces of attraction or repulsion between the molecules. 3) The molecules are in constant, rapid, random motion and collide with one another and with the walls of the container. 4) The collisions are perfectly elastic (no loss of kinetic energy). 5) The average kinetic energy of the molecules is directly proportional to the absolute temperature.
KE(avg) proportional to T
Marking-scheme points
- ✓Molecular volume negligible; no intermolecular forces
- ✓Constant random motion, perfectly elastic collisions
- ✓Average KE proportional to absolute temperature
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States of Matter
Write the van der Waals equation for n moles of a real gas and explain the significance of the constants a and b.
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The van der Waals equation is (P + a n^2/V^2)(V - nb) = nRT. The constant 'a' corrects for the intermolecular forces of attraction (it accounts for the pressure being lower than ideal), and the constant 'b' corrects for the finite volume actually occupied by the gas molecules (excluded volume). Real gases deviate from ideal behaviour at high pressure and low temperature.
(P + a n^2/V^2)(V - nb) = nRT
Marking-scheme points
- ✓(P + a n^2/V^2)(V - nb) = nRT
- ✓a: correction for intermolecular attraction
- ✓b: correction for finite molecular volume
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Thermodynamics
Define system and surroundings. Name the three types of thermodynamic systems.
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A system is the specified part of the universe under study; the surroundings are the rest of the universe outside the system that can interact with it. The three types are: open system (exchanges both matter and energy with surroundings), closed system (exchanges only energy, not matter) and isolated system (exchanges neither matter nor energy).
Marking-scheme points
- ✓System = part under study; surroundings = rest of universe
- ✓Open: exchanges matter and energy
- ✓Closed: only energy; Isolated: neither
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Thermodynamics
State the first law of thermodynamics and give its mathematical expression with sign convention.
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The first law of thermodynamics states that energy can neither be created nor destroyed, only transformed from one form to another; the total energy of an isolated system remains constant. Mathematically, delta U = q + w, where delta U is the change in internal energy, q is the heat added to the system (positive when absorbed) and w is the work done on the system (positive when done on the system).
delta U = q + w
Marking-scheme points
- ✓Energy is conserved (cannot be created or destroyed)
- ✓delta U = q + w
- ✓q positive if heat absorbed; w positive if work done on system
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Thermodynamics
When 1 kJ of heat is supplied to a gas, it does 200 J of work by expanding. Calculate the change in internal energy of the gas.
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Heat supplied to the system q = +1 kJ = +1000 J. Work is done BY the gas, so work done on the system w = -200 J. By the first law, delta U = q + w = 1000 + (-200) = 800 J. The internal energy increases by 800 J.
delta U = q + w
Marking-scheme points
- ✓q = +1000 J (heat absorbed)
- ✓Work done by gas -> w = -200 J
- ✓delta U = q + w = 1000 - 200 = 800 J
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Thermodynamics
Define enthalpy. Derive the relation between delta H and delta U for a reaction involving gases.
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Enthalpy (H) is the total heat content of a system at constant pressure, defined as H = U + PV. For a reaction at constant pressure and temperature, delta H = delta U + P delta V. For ideal gases, P delta V = delta ng RT, where delta ng is the change in the number of moles of gaseous species. Hence delta H = delta U + delta ng RT.
delta H = delta U + delta ng RT
Marking-scheme points
- ✓H = U + PV (heat content at constant pressure)
- ✓delta H = delta U + P delta V
- ✓For gases: delta H = delta U + delta ng RT
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Thermodynamics
Given: C(s) + O2(g) -> CO2(g), delta H = -393.5 kJ and CO(g) + 1/2 O2(g) -> CO2(g), delta H = -283.0 kJ. Calculate the enthalpy of formation of CO(g).
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We want C(s) + 1/2 O2(g) -> CO(g). By Hess's law, subtract the second equation from the first: delta H(reqd) = delta H1 - delta H2 = (-393.5) - (-283.0) = -110.5 kJ. So the enthalpy of formation of CO is -110.5 kJ/mol.
delta H(reqd) = delta H1 - delta H2
Marking-scheme points
- ✓Target: C + 1/2 O2 -> CO
- ✓Subtract equation 2 from equation 1
- ✓delta H = -393.5 - (-283.0) = -110.5 kJ/mol
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Thermodynamics
State Hess's law of constant heat summation and mention one of its applications.
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Hess's law states that the total enthalpy change of a reaction is the same whether the reaction takes place in one step or in several steps, provided the initial and final conditions are the same. It follows from the fact that enthalpy is a state function. Applications: it is used to calculate enthalpies of formation, bond enthalpies and reaction enthalpies that cannot be measured directly.
Marking-scheme points
- ✓Total enthalpy change is path independent
- ✓Consequence of enthalpy being a state function
- ✓Used to find delta H that cannot be measured directly
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Thermodynamics
What is entropy? Predict the sign of delta S when ice melts into water.
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Entropy (S) is a thermodynamic state function that measures the degree of randomness or disorder of a system. When ice (a highly ordered solid) melts into water (a more disordered liquid), disorder increases, so the entropy increases and delta S is positive.
delta S = q(rev) / T
Marking-scheme points
- ✓Entropy = measure of randomness/disorder
- ✓Melting increases disorder (solid -> liquid)
- ✓delta S is positive for melting of ice
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Thermodynamics
Write the Gibbs-Helmholtz equation and state the criteria of spontaneity in terms of delta G.
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The Gibbs energy change is given by delta G = delta H - T delta S. Criteria of spontaneity: if delta G < 0 (negative), the process is spontaneous; if delta G = 0, the system is at equilibrium; if delta G > 0 (positive), the process is non-spontaneous (the reverse is spontaneous). A reaction is always spontaneous when delta H is negative and delta S is positive.
delta G = delta H - T delta S
Marking-scheme points
- ✓delta G = delta H - T delta S
- ✓delta G < 0: spontaneous; = 0: equilibrium; > 0: non-spontaneous
- ✓delta H negative and delta S positive -> always spontaneous
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Thermodynamics
For a reaction, delta H = -92.4 kJ and delta S = -198 J/K at 298 K. Calculate delta G and predict whether the reaction is spontaneous.
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Convert delta S to kJ: -198 J/K = -0.198 kJ/K. delta G = delta H - T delta S = -92.4 - (298 x -0.198) = -92.4 - (-59.0) = -92.4 + 59.0 = -33.4 kJ. Since delta G is negative, the reaction is spontaneous at 298 K.
delta G = delta H - T delta S
Marking-scheme points
- ✓delta G = delta H - T delta S
- ✓T delta S = 298 x (-0.198) = -59.0 kJ
- ✓delta G = -33.4 kJ -> spontaneous
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Thermodynamics
Define standard enthalpy of formation and standard enthalpy of combustion.
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Standard enthalpy of formation is the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states (at 298 K and 1 bar); e.g. for CO2 it is -393.5 kJ/mol. Standard enthalpy of combustion is the enthalpy change when one mole of a substance is completely burnt in excess oxygen under standard conditions; it is always negative (exothermic).
Marking-scheme points
- ✓Formation: 1 mol compound from elements in standard states
- ✓Combustion: 1 mol substance completely burnt in oxygen
- ✓Enthalpy of combustion is always negative
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Equilibrium
What is a reversible reaction? State two characteristics of chemical equilibrium.
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A reversible reaction is one that proceeds in both forward and backward directions under the same conditions. Characteristics of chemical equilibrium: (1) it is dynamic in nature, i.e. the forward and backward reactions continue at equal rates; (2) the observable properties (concentration, pressure, colour) remain constant with time; (3) it can be attained from either direction and is disturbed by changing conditions.
rate(forward) = rate(backward)
Marking-scheme points
- ✓Reversible: proceeds in both directions
- ✓Equilibrium is dynamic: forward rate = backward rate
- ✓Measurable properties stay constant with time
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Equilibrium
For the reaction N2(g) + 3H2(g) <=> 2NH3(g), write the expression for Kc and state the relation between Kp and Kc.
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Kc = [NH3]^2 / ([N2][H2]^3). The relation between Kp and Kc is Kp = Kc (RT)^(delta ng), where delta ng = (moles of gaseous products) - (moles of gaseous reactants). Here delta ng = 2 - (1 + 3) = -2, so Kp = Kc (RT)^-2.
Kp = Kc (RT)^(delta ng)
Marking-scheme points
- ✓Kc = [NH3]^2 / ([N2][H2]^3)
- ✓Kp = Kc (RT)^(delta ng)
- ✓delta ng = 2 - 4 = -2, so Kp = Kc (RT)^-2
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Equilibrium
For the equilibrium N2O4(g) <=> 2NO2(g), the equilibrium concentrations are [N2O4] = 0.02 mol/L and [NO2] = 0.04 mol/L. Calculate Kc.
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Kc = [NO2]^2 / [N2O4] = (0.04)^2 / 0.02 = 0.0016 / 0.02 = 0.08 mol/L. The units are mol/L because delta ng = 1 for this reaction.
Kc = [products]^coeff / [reactants]^coeff
Marking-scheme points
- ✓Kc = [NO2]^2 / [N2O4]
- ✓= (0.04)^2 / 0.02 = 0.0016/0.02
- ✓Kc = 0.08 mol/L
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Equilibrium
State Le Chatelier's principle. What is the effect of increasing pressure on the equilibrium N2(g) + 3H2(g) <=> 2NH3(g)?
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Le Chatelier's principle states that if a system at equilibrium is subjected to a change in concentration, pressure or temperature, the equilibrium shifts in the direction that tends to counteract (reduce) the effect of that change. Increasing the pressure shifts this equilibrium in the forward direction (towards NH3), because the forward reaction reduces the number of gas moles from 4 (1 + 3) to 2, thereby lowering the pressure.
Marking-scheme points
- ✓System shifts to oppose the imposed change
- ✓Higher pressure favours the side with fewer gas moles
- ✓Here forward reaction (4 -> 2 moles) is favoured -> more NH3
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Equilibrium
What is the effect of temperature and of a catalyst on a system at equilibrium?
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Temperature: increasing temperature favours the endothermic direction and decreasing temperature favours the exothermic direction (as per Le Chatelier's principle); temperature also changes the value of the equilibrium constant K. Catalyst: a catalyst speeds up both the forward and backward reactions equally, so it helps the system reach equilibrium faster but does not shift the position of equilibrium or change the value of K.
Marking-scheme points
- ✓Higher temperature favours endothermic direction; changes K
- ✓Catalyst speeds up forward and backward reactions equally
- ✓Catalyst does not shift equilibrium or change K
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