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The questions your board exam loves to ask

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.

800 board questions•Model answers•Marking-scheme points•Every chapter•CBSE · ISC · State boards

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ChemistryClass 113 marksmedium

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.

Reveal model answer + marking points▾

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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ChemistryClass 113 markshard

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).

Reveal model answer + marking points▾

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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ChemistryClass 113 marksmedium

Thermodynamics

Write the Gibbs-Helmholtz equation and state the criteria of spontaneity in terms of delta G.

Reveal model answer + marking points▾

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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