Chemical Thermodynamics
NIOS · Class 12 · Chemistry
Most important questions from Chemical Thermodynamics for NIOS Class 12 Chemistry board exam 2026. MCQs, short answer, and long answer questions with marks.
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For the combustion of ethanol: C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l); ΔH = −1366 kJ/mol. How much heat is released on combustion of 23 g of ethanol? (Molar mass of C₂H₅OH = 46 g/mol)
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683 kJ
Step 1: Find moles of ethanol = mass/molar mass = 23/46 = 0.5 mol. Step 2: The given ΔH = −1366 kJ refers to combustion of 1 mole of ethanol. Step 3: Heat released for 0.5 mol = 0.5 × 1366 = 683 kJ. Step 4: The sign is negative (exothermic), so heat RELEASED = 683 kJ. Option 1366 kJ is for 1 mole; 2732 kJ would be for 2 moles; 341.5 kJ would be the answer if 23 g were taken as 1/4 mol, which is incorrect.
Which of the following processes is an example of an ENDOTHERMIC reaction?
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Dissolving NH₄Cl in water (test tube feels cold)
Step 1: Endothermic reactions ABSORB heat from surroundings, making the surroundings (or the test tube) feel cold. Step 2: Dissolving NH₄Cl in water absorbs heat → test tube feels cold → endothermic. Step 3: Combustion of coal releases heat and light → exothermic. Step 4: Zinc + HCl produces H₂ gas and the test tube becomes warm → exothermic. Adding water to CaO releases a lot of heat → highly exothermic. The key observation for endothermic reaction is cooling of surroundings.
According to Hess's Law, if C(graphite) + O₂(g) → CO₂(g); ΔH₁ = −394 kJ and CO(g) + ½O₂(g) → CO₂(g); ΔH₂ = −283 kJ, then ΔH for C(graphite) + ½O₂(g) → CO(g) is:
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−111 kJ/mol
Step 1: Write the target reaction: C(graphite) + ½O₂(g) → CO(g); ΔH = ? Step 2: According to Hess's Law, ΔH = ΔH₁ − ΔH₂ (subtract reaction 2 from reaction 1). Step 3: ΔH = (−394) − (−283) = −394 + 283 = −111 kJ/mol. Step 4: This works because CO₂ appears on the right in both equations; subtracting equation 2 reverses it and cancels CO₂ and ½O₂. The positive value +111 kJ arises from a sign error. −677 kJ comes from incorrectly adding both values. Hess's Law is valid because enthalpy is a state function.
The standard enthalpy of formation (ΔfH°) of an element in its most stable state is:
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Zero
Step 1: Standard enthalpy of formation is defined as the enthalpy change when ONE mole of a compound is formed from its elements in their most stable states. Step 2: By convention, since elements in their most stable form are the reference state, no energy change occurs in 'forming' them from themselves. Step 3: For example, ΔfH°(O₂, g) = 0, ΔfH°(C, graphite) = 0, ΔfH°(Fe, s) = 0. Step 4: This is a universally accepted convention used as the reference for all thermochemical calculations. It does NOT mean elements have no energy; it simply means we set their enthalpy as the baseline (zero refer
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