The enthalpy of S-S bond from the following data is- i C2H5-S-C2H5g- - fHo-147-2 kJ - mol ii C2H5-S-S-C2H5g- - fHo-201-9 kJ - mol iii Sg- - fHo-222-8 kJ - mol

The correct option is D 277.5 kJ / molGiven; (i) 4C(s)+5 H _ 2 ( g ) +S(s)→ C _ 2 H _ 5 S C _ 2 H _ 5 ( g ) #160; #160; Δ #160;_ f ...

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Byju's Answer

Standard XII

Chemistry

Heat of Formation

The enthalpy ...

Question

The enthalpy of $S - S$ bond from the following data is:

( i ) $C_{2} H_{5} - S - C_{2} H_{5} (g); Δ_{f} H^{o} = - 147.2$ kJ / mol

( ii ) $C_{2} H_{5} - S - S - C_{2} H_{5} (g); Δ_{f} H^{o} = - 201.9$ kJ / mol

( iii ) $S (g); Δ_{f} H^{o} = - 222.8$ kJ / mol

124.5 kJ / mol

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887.5 kJ / mol

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967.5 kJ / mol

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277.5 kJ / mol

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Solution

The correct option is D 277.5 kJ / mol
Given;
$(i)$ $4 C (s) + 5 H_{2} (g) + S (s) \to C_{2} H_{5} - S - C_{2} H_{5} (g)$ $Δ_{f} H = - 147.23 k J / m o l$
$(i i)$ $4 C (s) + 5 H_{2} (g) + 2 S (s) \to C_{2} H_{5} - S - S - C_{2} H_{5} (g)$ $Δ_{f} H = - 201.92 k J / m o l$
$(i i) - (i)$ gives
$C_{2} H_{5} - S - C_{2} H_{5} (g) + S (s) \to C_{2} H_{5} - S - S - C_{2} H_{5} (g)$
$Δ_{f} H = - 201.92 + 147.23 = - 54.69 k J / m o l$
Add the above equation to the following equation
$S (g) \to S (s)$ $Δ_{f} H = - 222.80 k J / m o l$ to get,
$C_{2} H_{5} - S - C_{2} H_{5} (g) + S (g) \to C_{2} H_{5} - S - S - C_{2} H_{5} (g)$
$Δ_{f} H = - 222.80 + (- 54.69) = - 277.49 k J / m o l$
This heat is evolved because of formation of $S - S$ bond.
So, bond enthapy of $S - S$ is $277.49 k J / m o l$

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The enthalpy of S−S bond from the following data is:( i ) C2H5−S−C2H5(g);ΔfHo=−147.2 kJ / mol( ii ) C2H5−S−S−C2H5(g);ΔfHo=−201.9 kJ / mol( iii ) S(g);ΔfHo=−222.8 kJ / mol

The enthalpy of $S - S$ bond from the following data is:

( i ) $C_{2} H_{5} - S - C_{2} H_{5} (g); Δ_{f} H^{o} = - 147.2$ kJ / mol

( ii ) $C_{2} H_{5} - S - S - C_{2} H_{5} (g); Δ_{f} H^{o} = - 201.9$ kJ / mol

( iii ) $S (g); Δ_{f} H^{o} = - 222.8$ kJ / mol