For the given reaction- H 2 g - Cl 2 g - 2 H - aq -2 Cl - aq - - G 0 -262-4kJThe value of free energy of formation - G f 0 for the ion Cl -1 aq - therefore will be-

The correct option is A 131.2 kJ mol ^ 1 ( Δ G ^ 0 ) reaction #160;=Δ G _ f ^ 0 (Products) Δ G _ f ^ 0 (reactants) ∴ ...

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Standard XII

Chemistry

Gibbs Free Energy & Spontaneity

For the given...

Question

For the given reaction,

$H_{2} (g) + {C l}_{2} (g) ⟶ 2 H^{+} (a q) + 2 {C l}^{-} (a q)$ ;

$Δ G^{0} = - 262.4 k J$

The value of free energy of formation $(Δ G_{f}^{0})$ for the ion ${C l}^{- 1} (a q)$ , therefore will be:

- 131.2

k J

{m o l}^{- 1}

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+ 131.2

k J

{m o l}^{- 1}

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- 262.4

k J

{m o l}^{- 1}

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+ 262.4

k J

{m o l}^{- 1}

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Solution

The correct option is A $- 131.2$ $k J$ ${m o l}^{- 1}$
$(Δ G^{0})$ reaction $= Δ G_{f}^{0}$ (Products) $- Δ G_{f}^{0}$ (reactants)
$∴ 264.4 = [2 Δ G_{f}^{0} (H^{+}) + 2 Δ G_{f}^{0} (C^{1 -})]$
or $264.4 = - [Δ G_{f}^{0} (H_{2}) + Δ G_{f}^{0} ({C l}_{2})]$
$= [0 + 2 Δ G_{f}^{0} ({C l}^{-})] + [0 + 0]$
or, $- 262.4 = 2 Δ G_{f}^{0} ({C l}^{-})$
or, $Δ G_{f}^{0} ({C l}^{-}) = - 131.2$ $k J$ ${m o l}^{- 1}$

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Similar questions

Q. The standard free energy of formation for

A g C l

298 K

- 109.7 k J

{m o l}^{- 1}

Δ G^{o} ({A g}^{+}) = 77.2 k J / m o l

;

Δ G^{o} ({C l}^{-}) = - 131.2 k J / m o l

. Find the solubility of

A g C l

0.05 M

K C l

. Neglect any complication due to complexation.

Q. Calculate the lattice enthalpy of

C a C l_{2}

, given that :
Enthalpy of sublimation for

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

Enthalpy of dissociation of

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

Ionisation energy of

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

Electron gain enthalpy of

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

Enthalpy of formation of

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

Q. Consider the following reactions.

(a)

H_{(a q)}^{+} + O H_{(a q)}^{-} = H_{2} O_{(l)}, Δ H = - X_{1}

m o l^{- 1}

(b)

H_{2 (g)} + \frac{1}{2} O_{2 (g)} = H_{2} O_{(l)}, Δ H = X_{2}

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(c)

C O_{2 (g)} + H_{2 (g)} = C O_{(g)} + H_{2} O_{(l)} - X_{3}

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C_{2} H_{2 (g)} + \frac{5}{2} O_{2 (g)} = 2 C O_{2 (g)} + H_{2} O_{(l)} + X_{4}

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Enthalpy of formation of

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Q. What will be

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

?
(B.E. of C - H and C - Cl bonds are 416 kJ

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and 325 kJ

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respectively)

Q. For the reaction

N_{2} + 3 X_{2} \to 2 N X_{3}

where

X = F, C l

(the average bond energies are

F - F = 155 k J m o l^{- 1}

N - F = 272 k J m o l^{- 1}

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and

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For the given reaction,H2(g)+Cl2(g)⟶2H+(aq)+2Cl−(aq);ΔG0=−262.4kJThe value of free energy of formation (ΔG0f) for the ion Cl−1(aq), therefore will be:

The correct option is A −131.2 kJ mol−1(ΔG0) reaction =ΔG0f (Products) −ΔG0f (reactants)∴264.4=[2ΔG0f(H+)+2ΔG0f(C1−)]or 264.4=−[ΔG0f(H2)+ΔG0f(Cl2)] =[0+2ΔG0f(Cl−)]+[0+0]or, −262.4=2ΔG0f(Cl−)or, ΔG0f(Cl−)=−131.2 kJ mol−1

For the given reaction,

$H_{2} (g) + {C l}_{2} (g) ⟶ 2 H^{+} (a q) + 2 {C l}^{-} (a q)$ ;

$Δ G^{0} = - 262.4 k J$

The value of free energy of formation $(Δ G_{f}^{0})$ for the ion ${C l}^{- 1} (a q)$ , therefore will be: