Find the enthalpy of S-S bond from the following data only magnitude in nearest integer in kJ-mol- i C2H5-S-C2H5g -Hof-147-2kJ-mol ii C2H5-S-S-C2H5g -Hof-201-9kJ-mol iii Sg -Hof-222-8kJ-mol

C_2H_5 S C_2H_5(g) + S(g) → #160;C_2H_5 S S C_2H_5 (g); Δ H The above reaction involves #160;1. Breaking S C bond (x amount of #160;energy absorbed ...

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Byju's Answer

Standard XII

Chemistry

Hess' Law

Find the enth...

Question

Find the enthalpy of $S - S$ bond from the following data (only magnitude in nearest integer in kJ/mol):

$(i) C_{2} H_{5} - S - C_{2} H_{5} (g)$ $Δ H_{f}^{o} = - 147.2 k J / m o l$
$(i i) C_{2} H_{5} - S - S - C_{2} H_{5} (g)$ $Δ H_{f}^{o} = - 201.9 k J / m o l$
$(i i i) S (g)$ $Δ H_{f}^{o} = 222.8 k J / m o l$

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Solution

$C_{2} H_{5} - S - C_{2} H_{5} (g) + S (g) \to C_{2} H_{5} - S - S - C_{2} H_{5} (g); Δ H$
The above reaction involves
1. Breaking S-C bond (x amount of energy absorbed)
2.Making S-S bond (y amount of energy released)
3. Making S-C bond (x amount of energy released)
Hence enthaply of the reaction $=$ x-y-x $=$ -y
And -y is nothing but the enthaply of $S - S$ bond
Therefore
Enthaply of $S - S$ bond $= Δ H = Σ Δ H_{f, p r o d u c t s}^{o} - Σ Δ H_{f, r e a c t a n t s}^{o} = - 201.9 + 147.2 + 222.8 = 168.1 k J / m o l$
= 168kJ \mol

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Q. Find the bond enthalpy of

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Find the enthalpy of S−S bond from the following data (only magnitude in nearest integer in kJ/mol):(i)C2H5−S−C2H5(g) ΔHof=−147.2kJ/mol(ii)C2H5−S−S−C2H5(g) ΔHof=−201.9kJ/mol(iii)S(g) ΔHof=222.8kJ/mol