Iv'e got a long one

The equilibrium constant for the reaction AgBr(s) Picture Ag+(aq) + Br− (aq) is the solubility product constant, Ksp = 7.7 × 10−

barxatty [35]

Answer:

The reaction will be non spontaneous at these concentrations.

Explanation:

$AgBr(s)\rightarrow Ag^+(aq) + Br^- (aq)$

Expression for an equilibrium constant $K_c$ :

$K_c=\frac{[Ag^+][Br^-]}{[AgCl]}=\frac{[Ag^+][Br^-]}{1}=[Ag^+][Br^-]$

Solubility product of the reaction:

$K_{sp}=[Ag^+][Br^-]=K_c=7.7\times 10^{-13}$

Reaction between Gibb's free energy and equilibrium constant if given as:

$\Delta G^o=-2.303\times R\times T\times \log K_c$

$\Delta G^o=-2.303\times R\times T\times \log K_{sp}$

$\Delta G^o=-2.303\times 8.314 J/K mol\times 298 K\times \log[7.7\times 10^{-13}]$

$\Delta G^o=69,117.84 J/mol=69.117 kJ/mol$

Gibb's free energy when concentration $[Ag^+] = 1.0\times 10^{-2} M$ and $[Br^-] = 1.0\times 10^{-3} M$

Reaction quotient of an equilibrium = Q

$Q=[Ag^+][Br^-]=1.0\times 10^{-2} M\times 1.0\times 10^{-3} M=1.0\times 10^{-5}$

$\Delta G=\Delta G^o+(2.303\times R\times T\times \log Q)$

$\Delta G=69.117 kJ/mol+(2.303\times 8.314 Joule/mol K\times 298 K\times \log[1.0\times 10^{-5}])$

$\Delta G=40.588 kJ/mol$

For reaction to spontaneous reaction: $\Delta G$ .
For reaction to non spontaneous reaction: $\Delta G>0$ .

Since ,the value of Gibbs free energy is greater than zero which means reaction will be non spontaneous at these concentrations

5 0

3 years ago

On the basis of the Ksp values below, what is the order of the solubility from least soluble to most soluble for these compounds

Viktor [21]

Answer:

The order of solubility is AgBr < Ag₂CO₃ < AgCl

Explanation:

The solubility constant give us the molar solubilty of ionic compounds. In general for a compound AB the ksp will be given by:

Ksp = (A) (B) where A and B are the molar solubilities = s² (for compounds with 1:1 ratio).

It follows then that the higher the value of Ksp the greater solubilty of the compound if we are comparing compounds with the same ionic ratios:

Comparing AgBr: Ksp = 5.4 x 10⁻¹³ with AgCl: Ksp = 1.8 x 10⁻¹⁰, AgCl will be more soluble.

Comparing Ag2CO3: Ksp = 8.0 x 10⁻¹² with AgCl Ksp = AgCl: Ksp = 1.8 x 10⁻¹⁰ we have the complication of the ratio of ions 2:1 in Ag2CO3, so the answer is not obvious. But since we know that

Ag2CO3 ⇄ 2 Ag⁺ + CO₃²₋

Ksp Ag2CO3 = 2s x s = 2 s² = 8.0 x 10-12

s = 4 x 10⁻12 ∴ s= 2 x 10⁻⁶

And for AgCl

AgCl ⇄ Ag⁺ + Cl⁻

Ksp = s² = 1.8 x 10⁻¹⁰ ∴ s = √ 1.8 x 10⁻¹⁰ = 1.3 x 10⁻⁵

Therefore, AgCl is more soluble than Ag₂CO₃

The order of solubility is AgBr < Ag₂CO₃ < AgCl

8 0

3 years ago

Read 2 more answers

The specific heat of aluminum is 0.897 j/g•°c. which equation would you use to calculate the amount of heat needed to raise the

Varvara68 [4.7K]

7 0

3 years ago

32. Barium sulfate, BaSO4, is a white crystalline solid that is insoluble in water. It is used by doctors to diagnose

DIA [1.3K]

Answer:

42

Explanation:

bygyytrdesf

7 0

2 years ago

A gas in a sealed container had its volume increased from 12.1 liters to 21.1 liters.

ArbitrLikvidat [17]

Answer:

The answer to your question is P2 = 170.9 torr

Explanation:

Data

Volume 1 = 12.1 l Volume 2 = 21.1 l

Temperature 1 = 241 °K Temperature 2 = 298°K

Pressure 1 = 546 torr Pressure 2 = ?

Process

To solve this problem use the combined gas law.

P1V1/T1 = P2V2/T2

-Solve for P2

P2 = T1V1T2 / T1V2

-Substitution

P2 = (241 x 12.1 x 298) / (241 x 21.1)

-Simplification

P2 = 868997.8 / 5085.1

-Result

P2 = 170.9 torr

8 0

3 years ago