JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Chemical Thermodynamics

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The average S-F bond energy in kJ mol⁻¹ of SF₆ is ______ (Rounded off to the nearest integer) [Given : The values of standard enthalpy of formation of SF₆(g), S(g) and F(g) are -1100, 275 and 80 kJ mol⁻¹ respectively.]

Answer: 309

Solution

The average bond enthalpy is obtained from the formation of the molecule out of free gaseous atoms:

S (g) + 6 F (g) \to S F_{6} (g)

Its enthalpy change follows from the standard enthalpies of formation,

Δ H = Δ H_{f} (S F_{6}, g) - Δ H_{f} (S, g) - 6 Δ H_{f} (F, g) = - 1100 - 275 - 6 (80) = - 1855 kJ mo l^{- 1}

Six

S - F

bonds are made in this step and no bond is broken, so the average

S - F

bond energy is

\frac{1855}{6} = 309.2

, that is

309

kJ mo l^{- 1}

to the nearest integer.

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Integer answer

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