Given the following thermochemical data at 298 K and 1 barΔHvap ∘CH3OH=40.0kJmol−1ΔfH∘:H(g)=220kJmol−1O(g)=250kJmol−1C(g)=710kJmol−1Bond dissociation energy(C−H)=420kJmol−1(C−O)=350kJmol−1(O−H)=465kJmol−1 The ΔfH∘ of liquid methyl alcohol in kJmol−1 is

ΔfH∘=ΔfH∘(C,g)+4ΔfH∘(H,g)+ΔfH∘(O,g)−[3(BE)C−H+(BE)C−O+(BE)O−H=(710+4×220+250)−(30×420+350+465)=−235kJCH3OH(g)→CH3OH(l);ΔH∘=−40kJ Thus, ΔfH∘CH3OH,l=−235−40=−275kJmol−1 Note ΔfH∘, based on (BE) value requires all species in gaseous state thus, 40 kJ of heat is released which converted CH3OH (g) into CH3OH(l).

Given the following thermochemical data at 298 K and 1 barΔHvap ∘CH3OH=40.0kJmol−1ΔfH∘:H(g)=220kJmol−1O(g)=250kJmol−1C(g)=710kJmol−1Bond dissociation energy(C−H)=420kJmol−1(C−O)=350kJmol−1(O−H)=465kJmol−1 The ΔfH∘ of liquid methyl alcohol in kJmol−1 is

ΔfH∘=ΔfH∘(C,g)+4ΔfH∘(H,g)+ΔfH∘(O,g)−[3(BE)C−H+(BE)C−O+(BE)O−H=(710+4×220+250)−(30×420+350+465)=−235kJCH3OH(g)→CH3OH(l);ΔH∘=−40kJ Thus, ΔfH∘CH3OH,l=−235−40=−275kJmol−1 Note ΔfH∘, based on (BE) value requires all species in gaseous state thus, 40 kJ of heat is released which converted CH3OH (g) into CH3OH(l).

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Given the following thermochemical data at 298 K and 1 bar
$\begin{array}{l} {ΔH}_{vap}^{\circ} ({CH}_{3} OH) = 40.0 kJ {mol}^{- 1} \\ Δ_{f} H^{\circ} : H (g) = 220 kJ {mol}^{- 1} \\ O (g) = 250 kJ {mol}^{- 1} \\ C (g) = 710 kJ {mol}^{- 1} \end{array}$
Bond dissociation energy
$\begin{aligned} (C - H) = 420 kJ {mol}^{- 1} \\ (C - O) = 350 kJ {mol}^{- 1} \\ (O - H) = 465 kJ {mol}^{- 1} \end{aligned}$
$The Δ_{f} H^{\circ} of liquid methyl alcohol in kJ {mol}^{- 1} is$

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- 266 kJ mol^-1

-170 kJ mol^-1

+ 275 kJ mol^-1

+170 kJ mol^-1

answer is B.

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Detailed Solution

$\begin{array}{l} Δ_{f} H^{\circ} = Δ_{f} H^{\circ} (C, g) + 4 Δ_{f} H^{\circ} (H, g) + Δ_{f} H^{\circ} (O, g) - [3 (BE) \\ C - H + (BE)_{C - O} + (BE)_{O - H}] \\ = (710 + 4 \times 220 + 250) - (30 \times 420 + 350 + 465) = - 235 kJ \\ {CH}_{3} OH (g) \to {CH}_{3} OH (l); {ΔH}^{\circ} = - 40 kJ \end{array}$

$Thus, Δ_{f} H^{\circ} [{CH}_{3} OH, l] = - 235 - 40 = - 275 kJ {mol}^{- 1}$

$Note Δ_{f} H^{\circ}$ , based on (BE) value requires all species in gaseous state thus, 40 kJ of heat is released which converted CH₃OH (g) into CH₃OH( $l$ ).

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