At 298.2 K the relationship between enthalpy of bond dissociation (in kJ mol?¹) for hydrogen (E?) and its isotope, deuterium (E?), is best described by:
At 298.2 K the relationship between enthalpy of bond dissociation (in kJ mol?¹) for hydrogen (E?) and its isotope, deuterium (E?), is best described by:
Option 1 - <p>Eʜ = Eᴅ<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--></p>
Option 2 - <p>Eʜ = ½ Eᴅ<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--></p>
Option 3 - <p>Eʜ = 2Eᴅ<br><!-- [if !supportLineBreakNewLine]--><br><!--[endif]--></p>
Option 4 - <p>Eʜ ≈ Eᴅ - 7.5</p>
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7 months ago
Correct Option - 4
Detailed Solution:
B.D.E H-H < B.D.E D-D
B.D.E H-H = 435.9 kJ / mole
B.D.E D-D = 443.4 kJ / mole
E H ≈ E D - 7.5
Similar Questions for you
CH3COOH + NaOH → CH3COONa + H2O
ΔH = –50.6 kJ/mol
NaOH + SA [HCl] → NaCl + H2O
ΔH = –55.9 kJ/mol
the value of ΔH for ionisation of CH3COOH
⇒ ΔH = +55.9 – 50.6
5.3 kJ/mol
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