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Standard XIII

Chemistry

Heat of Formation

The value of ...

Question

The value of standard enthalpy of formation of $S F_{6} (g), S (g) a n d F (g) a r e - 1100, 275, 80 k J / m o l$ respectively. The average bond energy of $S F_{6}$ (in kJ/mol) is
(Give your answer upto two decimal points)

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Solution

$S F_{6} (g) \to S (g) + 6 F (g)$
$Δ H_{f}^{o} = H_{f}^{o} p r o d u c t s - H_{f r e a c t a n t s}^{o}$
$Δ H_{f}^{o} = (275 + 80 \times 6) - (- 1100)$
$Δ H_{f}^{o} = 1855 k J / m o l$
The average bond energy in $S F_{6}$ will be = $\frac{1855 k J / m o l}{6} = 309.16 k J / m o l$

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Q. The value of standard enthalpy of formation of

S F_{6} (g), S (g) a n d F (g) a r e - 1100, 275, 80 k J / m o l

respectively. The average bond energy of

S F_{6}

(in kJ/mol) is
(Give your answer upto two decimal points)

Q. The average S–F bond energy in

k J m o l^{- 1}

S F_{6}

is______. (Rounded off to the nearest integer)
[Given : The values of standard enthalpy of formation of

S F_{6} (g), S (g) and F (g) are 1100, 275 and 80 k J m o l^{- 1}

respectively.]

The standard heat of formation values of $S F_{6} (g)$ , $S (g)$ and $F (g)$ are: $- 1100, 275$ and $80 K J m o l^{- 1}$ respectively. Then the average S - F bond energy in $S F_{6}$ is:

Q. The average

S - F

bond energy in

k J m o l^{- 1}

S F_{6}

is _____.(Rounded off to the nearest integer

[Given : The values of standard enthalpy of formation of

S F_{6} (g), S (g) a n d F (g)

are - 1100, 275 and 80 kJ

m o l^{- 1}

respectively]

The standard heat of formation values of $S F_{6} (g), S (g)$ and $F (g)$ are $- 1100, 275$ and $80$ J mol $^{- 1}$ respecitvely. The $S - F$ bond energy in $S F_{6}$ is :

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Heat of Formation

Standard XIII Chemistry

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The value of standard enthalpy of formation of SF6(g), S(g) and F(g)are−1100,275,80 kJ/mol respectively. The average bond energy of SF6 (in kJ/mol) is (Give your answer upto two decimal points)

SF6(g)→S(g)+6F(g) ΔHof=Hofproducts−Hofreactants ΔHof=(275+80×6)−(−1100) ΔHof=1855 kJ/mol The average bond energy in SF6 will be = 1855 kJ/mol6=309.16 kJ/mol

The value of standard enthalpy of formation of $S F_{6} (g), S (g) a n d F (g) a r e - 1100, 275, 80 k J / m o l$ respectively. The average bond energy of $S F_{6}$ (in kJ/mol) is
(Give your answer upto two decimal points)

$S F_{6} (g) \to S (g) + 6 F (g)$
$Δ H_{f}^{o} = H_{f}^{o} p r o d u c t s - H_{f r e a c t a n t s}^{o}$
$Δ H_{f}^{o} = (275 + 80 \times 6) - (- 1100)$
$Δ H_{f}^{o} = 1855 k J / m o l$
The average bond energy in $S F_{6}$ will be = $\frac{1855 k J / m o l}{6} = 309.16 k J / m o l$