Enthalpies for Different Types of Reactions - Bond Enthalpy

Calculate the enthalpy change for the process

\[\ce{CCl_4 (g) → C(g) + 4Cl(g)}\]

and calculate bond enthalpy of C–Cl in CCl_4(g).

Δ_vapH^θ (CCl₄) = 30.5 kJ mol^–1.

Δ_fH^θ (CCl₄) = –135.5 kJ mol^–1.

Δ_aH^θ (C) = 715.0 kJ mol^–1, where Δ_aH^θ is enthalpy of atomisation

Δ_aH^θ (Cl₂) = 242 kJ mol^–1

The enthalpy of atomisation for the reaction \[\ce{CH4(g) -> C(g) + 4H(g)}\] is 1665 kJ mol^–1. What is the bond energy of \[\ce{C – H}\] bond?

Consider the reactions given below. On the basis of these reactions find out which of the algebric relations given in options (i) to (iv) is correct?

(a) \[\ce{C (g) + 4 H (g) -> CH4 (g); ∆_rH = xkJ mol^{-1}}\]

(b) \[\ce{C (graphic) + 2H2 (g) -> CH4 (g); ∆_rH = ykJ mol^{-1}}\]

Use the following data to calculate \[\ce{∆_{lattice}H^Θ}\] for \[\ce{NaBr}\].

\[\ce{∆_{sub}H^Θ}\] for sodium metal = 108.4 kJ mol^–1

Ionization enthalpy of sodium = 496 kJ mol^–1

Electron gain enthalpy of bromine = – 325 kJ mol^–1

Bond dissociation enthalpy of bromine = 192 kJ mol^–1

\[\ce{∆_fH^Θ}\] for \[\ce{NaBr (s)}\] = – 325 kJ mol^–1 

One mole of acetone requires less heat to vapourise than 1 mol of water. Which of the two liquids has higher enthalpy of vapourisation?

The net enthalpy change of a reaction is the amount of energy required to break all the bonds in reactant molecules minus amount of energy required to form all the bonds in the product molecules. What will be the enthalpy change for the following reaction.

\[\ce{H2(g) + Br(g) -> 2HBr(g)}\]

Given that Bond energy of \[\ce{H2, Br2}\] and \[\ce{HBr}\] is 435 kJ mol^–1, 192 kJ mol^–1 and 368 kJ mol^–1 respectively.