If standard enthalpies of formation of CaCl (s) (hypothetical) and that of CaCl2(s) are − 188 J mol-1 and − 795 kJ mol−1 respectively, calculate the value of standard heat of reaction for the following disproportionation reaction 2CaCl(s) longrightarrow CaCl2(s) +Ca(s)

Question

If standard enthalpies of formation of CaCl (s) (hypothetical) and that of CaCl2(s) are − 188 J mol-1 and − 795 kJ mol−1 respectively, calculate the value of standard heat of reaction for the following disproportionation reaction 2CaCl(s) longrightarrow CaCl2(s) +Ca(s) (A)  − 607 kJ mol−1  (B)  + 607 kJ mol−1  (C)  − 419 kJ mol−1  (D)  + 419 kJ mol−1

Tardigrade Admin · Accepted Answer

In disproportionation reaction same species (here

C a^{2 +}

) is oxidised as well as reduced.

Standard heat of reaction,

Δ H^{\circ} = Δ H_{P}^{\circ} - Δ H_{R}^{\circ}

= Δ H_{F}^{\circ} (C a C l_{2}) + Δ H_{F}^{\circ} (C a) - 2Δ H_{F}^{\circ} (C a Cl)

= (- 795 + 0) - (- 2 \times 188)

= - 419 k J m o l^{- 1}

$- 607 k J m o l^{- 1}$

$+ 607 k J m o l^{- 1}$

$- 419 k J m o l^{- 1}$

$+ 419 k J m o l^{- 1}$

Q. If standard enthalpies of formation of CaCl(s) (hypothetical) and that of CaCl2​(s) are −188Jmol−1 and −795kJmol−1 respectively, calculate the value of standard heat of reaction for the following disproportionation reaction 2CaCl(s)⟶CaCl2(s)+Ca(s)

−607kJmol−1

+607kJmol−1

−419kJmol−1

+419kJmol−1

In disproportionation reaction same species (here Ca2+ ) is oxidised as well as reduced. Standard heat of reaction, ΔH∘=ΔHP∘​−ΔHR∘​ =ΔHF∘​(CaCl2​)+ΔHF∘​(Ca)−2ΔHF∘​(CaCl) =(−795+0)−(−2×188) =−419kJmol−1

$- 607 k J m o l^{- 1}$

$+ 607 k J m o l^{- 1}$

$- 419 k J m o l^{- 1}$

$+ 419 k J m o l^{- 1}$