**(a)** For the given reaction, we have

∆G° = ∆G°(Products) – ∆G°(Reactants)

∆G° = 52.0 – (87.0 + 0)

= -35.0 KJ mol^{-1}

**(b)** We know that,

∆G° = RT log K_{c}

∆G° = 2.303 RT log K_{c}

\( K_{c}=\dfrac{-35.0\times 10^{-3}}{-2.303\times 8.314\times 298}\\ \\ \)

\( =6.134\\ \\ \)

\( K_{c}=antilog(6.134)\\ \\ \)

\( =1.36\times 10^{6}\)

Therefore, the equilibrium constant for the given reaction K_{c} is

\( 1.36\times 10^{6}\)

Answered by Pragya Singh | 1 year agoThe concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen sulphide is 1.0 × 10^{–19} M. If 10 mL of this is added to 5 mL of 0.04M solution of the following: FeSO_{4}, MnCl_{2}, ZnCl_{2} and CdCl_{2} . in which of these solutions precipitation will take place?

What is the minimum volume of water required to dissolve 1g of calcium sulphate at 298 K? (For calcium sulphate, K_{sp} is 9.1 × 10^{–6}).

What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide? (For iron sulphide, Ksp = 6.3 × 10^{–18}).

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Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate? (For cupric iodate K_{sp} = 7.4 × 10^{–8} ).