You visited us 3 times! Enjoying our articles? Unlock Full Access!

Equilibrium Constant and Standard Free Energy Change

The standard ...

Question

The standard entropy change (in $J K^{- 1} {m o l}^{- 1}$ ) for the given reaction is:

$H_{2} (g) + {C l}_{2} (g) ⟶ 2 H C l (g)$ at $25^{o} C$ .

Express the value of the nearest integer of the standard entropy change in the form of $x + y$ if the integer is $x y .$
(Given $S^{o}$ for $H_{2}, {C l}_{2}$ and $H C l$ are $0.13, 0.22$ and $0.19 k J . K^{- 1} {m o l}^{- 1}$ respectively.)

Open in App

Solution

Entropy change for a reaction is given by,
$Δ S^{o} = \sum S_{p r o d u c t s}^{o} - \sum S_{r e a c t a n t s}^{o}$
$Δ S^{o} = 2 \times 0.19 - (0.13 + 0.22)$
$Δ S^{o} = 0.03 k J / (K . m o l)$
Hence,

$Δ S^{o} = 0.03 \times 1000 = 30 J / (K . m o l)$
So, $x + y = 3 + 0 = 3$ is the answer.

Suggest Corrections

Similar questions

S^{o}

H_{2}, {C l}_{2}

H C l

0.13, 0.22

0.19 {k J K}^{- 1}

{m o l}^{- 1}

H_{2} + {C l}_{2} \to 2 H C l

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

H_{2} (g) = 130.6, {C l}_{2} (g) = 233

H C l (g) = 186.7

298 K

1

H_{2} (g) + {C l}_{2} (g) \to 2 H C l (g)

H_{2} (g) + C l_{2} (g) ⟶ 2 H C l (g)

298 K

S_{H_{2}}^{⊖} = 131 J K^{- 1} m o l^{- 1}

S_{C l_{2}}^{⊖} = 223 J K^{- 1} m o l^{- 1}

S_{H C l}^{⊖} = 187 J K^{- 1} m o l^{- 1}

If S⁰ for H₂, Cl₂ and HCL are 0.13, 0.22 and 0.19 kJK^-1mol^-1 respectively. The total change in standard entropy for th2e reaction, H₂+Cl₂→2HCl

S_{H_{2}}^{\circ} = 131 J K^{- 1} m o l^{- 1} S_{C l_{2}}^{\circ} = 223 J K^{- 1} m o l^{- 1} a n d S_{H C l}^{\circ} = 187 J K^{- 1} m o l^{- 1}

1

H C l (g)

H_{2} (g)

C l_{2} (g)

H C l

H_{2} + C l_{2} \to 2 H C l

Join BYJU'S Learning Program