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Nuclear Stability

The standard ...

Question

The standard enthalpy of formation $△ H_{f}^{\circ}$ , of hypothetical CaCl(s) theoretically found to be $- 188 k J m o l^{- 1}$ and that of $C a C l_{2} (s)$ is $- 795 k J m o l^{- 1}$ .

Which of the following compounds is more stable?

C a C l (s)

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C a C l_{2} (s)

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Both are equally stable

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Nothing can be said

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Solution

The correct option is C $C a C l_{2} (s)$
As standard enthalpy of formation for $C a C l_{2}$ is more negative so it is more stable compound.

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Similar questions

C a C l (s)

C a {C l}_{2} (s)

- 188 J {m o l}^{- 1}

- 795 J {m o l}^{- 1}

2 C a C l (s) ⟶ C a {C l}_{2} (s) + C a (s)

{C a C l}_{2}

C a C l_{2}

C a (s) \to C a (g) = 121 k J m o l^{- 1}

C l_{2} (g) \to 2 C l (g) = 242.8 k J m o l^{- 1}

C a (g) \to C a^{+ +} (g) = 2422 k J m o l^{- 1}

2 C l \to 2 C l^{- 1} = (2 \times - 355) k J m o l^{- 1} = - 710 k J m o l^{- 1}

C a C l_{2} = - 795 k J m o l^{- 1}

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