The equilibrium constant at 25oC for the process- Co3-aq- - 6NH3aq- - -CoNH36-3- aq- is 2 - 107-Calculate the value of Go in kJ-mol at 25oC- R - 8-314 J K-1mol-1Report you answer upto two decimal places

We know that, G^o = 2.303RT log K_c Given, K_c = 2 × 10^7, T = 298K, R = 8.314 J K^ 1 mol^ 1 Thus, from above equation, nbsp; G^o = 2.303 × 8.314 × 298 log ...

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Standard XIII

Chemistry

Gibbs Free Energy & Spontaneity

The equilibri...

Question

The equilibrium constant at $25^{o} C$ for the process:
$C o^{3 +} (a q .) + 6 N H_{3} (a q .) ⇌ [C o (N H_{3})_{6}]^{3 +} (a q .) i s 2 \times 10^{7} .$
Calculate the value of $△ G^{o} (i n k J / m o l) a t 25^{o} C . (R = 8.314 J K^{- 1} m o l^{- 1})$
(Report you answer upto two decimal places)

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Solution

We know that, $△ G^{o} = - 2.303 R T l o g K_{c}$
Given, $K_{c} = 2 \times 10^{7}, T = 298 K, R = 8.314 J K^{- 1} m o l^{- 1}$
Thus, from above equation,
$△ G^{o} = - 2.303 \times 8.314 \times 298 l o g 2 \times 10^{7}$
$= - 41658.56 J / m o l = - 41.66 k J / m o l$

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Similar questions

Q. The equilibrium constant at

25^{o} C

for the process:

{C o}^{3 +} (a q .) + 6 {N H}_{3} (a q .) ⇌ {[C o {({N H}_{3})}_{6}]}^{3 +} (a q .)

2 \times 10^{7}

Calculate the value of

Δ G^{o}

25^{o} C

(R = 8.314 J K^{- 1} {m o l}^{- 1})

Q. The equilibrium constant at

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for the process:

C o^{3 +} (a q .) + 6 N H_{3} (a q .) ⇌ [C o (N H_{3})_{6}]^{3 +} (a q .) i s 2 \times 10^{7} .

Calculate the value of

△ G^{o} (i n k J / m o l) a t 25^{o} C . (R = 8.314 J K^{- 1} m o l^{- 1})

(Report you answer upto two decimal places)

Q. Calculate the equilibrium constant for the following reaction:

C u (s) + 2 {A g}^{+} (a q .) ⇌ {C u}^{2 +} (a q) + 2 A g (s)

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E_{c e l l}^{o} = 0.47 v o l t, R = 8.314 J K^{- 1} {m o l}^{- 1}, F = 96500 c o u l o m b

Q. Calculate the equilibrium constant for the reaction, at

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K_{p} = 1.7 \times 10^{7}

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The equilibrium constant at 25oC for the process: Co3+(aq.)+6NH3(aq.)⇌[Co(NH3)6]3+(aq.) is 2×107. Calculate the value of △Go (in kJ/mol) at 25oC. (R=8.314J K−1mol−1) (Report you answer upto two decimal places)

We know that, △Go=−2.303RT log Kc Given, Kc=2×107,T=298K,R=8.314 J K−1 mol−1 Thus, from above equation, △Go=−2.303×8.314×298 log 2×107 =−41658.56 J/mol=−41.66 kJ/mol

We know that, $△ G^{o} = - 2.303 R T l o g K_{c}$
Given, $K_{c} = 2 \times 10^{7}, T = 298 K, R = 8.314 J K^{- 1} m o l^{- 1}$
Thus, from above equation,
$△ G^{o} = - 2.303 \times 8.314 \times 298 l o g 2 \times 10^{7}$
$= - 41658.56 J / m o l = - 41.66 k J / m o l$