Calculate the magnitude of ring strain energy in ( kJ / mol ) of cyclopropane from the following data Δ f H [ C 3 H ( g )]=55 ; Δ f H [ CH ( g )]=715 ; Δ f H [ H ( g )]=220 ; BE ( C - C )=355 ; BE ( C - H )=410 (All in kJ / mol ):

Question

Calculate the magnitude of ring strain energy in ( kJ / mol ) of cyclopropane from the following data Δ f H [ C 3 H ( g )]=55 ; Δ f H [ CH ( g )]=715 ; Δ f H [ H ( g )]=220 ; BE ( C - C )=355 ; BE ( C - H )=410 (All in kJ / mol ): (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/85666cc99fbe875b77178093eec21872-.png" /> Δ H 4=- B.E. (3 C - C +6 C - H )+ Ring strain energy (RSE) =Δ H 1-Δ H 2-Δ H 3 R.S.E. =Δ H 1-Δ H 2-Δ H 3+ text B.E. (3 C - C +6 C - H ) =55-2145-1320+(3 × 355+6 × 410) =115 kJ / mol

Q. Calculate the magnitude of ring strain energy in $(k J / m o l)$ of cyclopropane from the following data
$Δ_{f} H [C_{3} H (g)] = 55; Δ_{f} H [C H (g)] = 715;$
$Δ_{f} H [H (g)] = 220; BE (C - C) = 355; BE (C - H) = 410$
(All in $k J / m o l$ ):

$Δ H_{4} = -$ B.E. $(3 C - C + 6 C - H) +$ Ring strain energy (RSE)
$= Δ H_{1} - Δ H_{2} - Δ H_{3}$
$R . S . E . = Δ H_{1} - Δ H_{2} - Δ H_{3} + B.E. (3 C - C + 6 C - H)$
$= 55 - 2145 - 1320 + (3 \times 355 + 6 \times 410)$
$= 115 k J / m o l$

Q. Calculate the magnitude of ring strain energy in (kJ/mol) of cyclopropane from the following data Δf​H[C3​H(g)]=55;Δf​H[CH(g)]=715; Δf​H[H(g)]=220;BE(C−C)=355;BE(C−H)=410 (All in kJ/mol ):

ΔH4​=− B.E. (3C−C+6C−H)+ Ring strain energy (RSE) =ΔH1​−ΔH2​−ΔH3​ R.S.E.=ΔH1​−ΔH2​−ΔH3​+ B.E. (3C−C+6C−H) =55−2145−1320+(3×355+6×410) =115kJ/mol

Q. Calculate the magnitude of ring strain energy in $(k J / m o l)$ of cyclopropane from the following data
$Δ_{f} H [C_{3} H (g)] = 55; Δ_{f} H [C H (g)] = 715;$
$Δ_{f} H [H (g)] = 220; BE (C - C) = 355; BE (C - H) = 410$
(All in $k J / m o l$ ):

$Δ H_{4} = -$ B.E. $(3 C - C + 6 C - H) +$ Ring strain energy (RSE)
$= Δ H_{1} - Δ H_{2} - Δ H_{3}$
$R . S . E . = Δ H_{1} - Δ H_{2} - Δ H_{3} + B.E. (3 C - C + 6 C - H)$
$= 55 - 2145 - 1320 + (3 \times 355 + 6 \times 410)$
$= 115 k J / m o l$