Chemical Thermodynamics — Important Questions

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.

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.

−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.

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