Chemistry, 14.07.2019 13:30 Yorlin4441

Use the thermodynamic data at 298 k below to determine the ksp for barium carbonate, baco3 at this temperature. substance: ba2+(aq) co32–(aq) baco3(s) δh°f (kj/mol): –538.36 –676.26 –1219 δg°f (kj/mol): –560.7 –528.1 –1139 s°(j/k·mol): 13 –53.1 112

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Answer from: 191600

Answer : the correct answer for ksp = 1.59 * 10⁻⁹

Following are the steps to calculate the ksp of reaction

BaCO₃ →Ba ²⁺ + CO₃²⁻ :

Step 1 : To find ΔG° of reaction :

ΔG° of reaction can be calculates by taking difference between ΔG° of products and reactants as :

ΔG° reaction =Sum of ΔG° ( products ) - Sum of Δ G° ( reactants ) .

Given : ΔG° for Ba²⁺ ( product )= -560.7 $\frac{KJ}{Mol}$

ΔG° for CO₃²⁻ (product ) =- 528.1 $\frac{KJ}{Mol}$

ΔG° BaCO₃ ( reactant) = –1139 $\frac{KJ}{Mol}$

Plugging value in formula :

ΔG° for reaction = ( ΔG° of Ba ²⁺ + ΔG° of CO₃²⁻ ) - (ΔG° of BaCO₃ )

⁻ = ( -560.7 $\frac{KJ}{Mol}$ + 528.1 $\frac{KJ}{Mol}$ ) - ( -1139 $\frac{KJ}{Mol}$ )

= ( -1088.8 $\frac{KJ}{Mol}$ ) - (-1139 $\frac{KJ}{Mol}$ )

= - 1088.8 $\frac{KJ}{Mol}$ + 1139 $\frac{KJ}{Mol}$

ΔG° of reaction = 50.2 $\frac{KJ}{Mol}$

Step 2: To calculate ksp from ΔG° of reaction .

The relation between Ksp and ΔG° is given as :

ΔG° = -RT ln ksp

Where ΔG° = Gibb's Free energy R = gas constant T = Temperature

Ksp = Solubility constant product .

Given : ΔG° of reaction = 50.2 $\frac{KJ}{Mol}$

T = 298 K R = 8.314 $\frac{J}{Mol * K}$

Plugging values in formula

$50.2 \frac{KJ}{mol} = - 8.314 \frac{J}{mol * K} * 298 K * ln ksp$

$50.2 \frac{KJ}{mol} = - 2477.572 \frac{J}{mol} * ln K$

((Converting 2477 $\frac{J}{mol} to \frac{KJ}{mol}$

Since , 1 KJ = 1000 J So , $2477 \frac{J}{mol} * \frac{1 KJ}{1000J} = 2.477 \frac{KJ}{mol}$ ))

Dividing both side by $- 2.477 \frac{KJ}{mol}$

$\frac{50.2\frac{KJ}{mol}}{-2.477 \frac{KJ}{mol}} = \frac{-2.477 \frac{KJ}{mol}}{-2.477 \frac{KJ}{mol}} * ln ksp$

ln ksp = $ln ksp = -20.27 \frac{KJ}{mol}$

Removing ln :

ksp = 1. 59 * 10⁻⁹

Answer from: Quest

vanderwaal's forces of attraction are the electrostatic intermolecular forces that exist in a molecule due to the temporary or permanent dipoles that are developed in a molecule. in $h_{2}s$ , the major intermolecular forces are dipole-dipole interactions as the molecule is polar (bent structure). so the vanderwaal's forces of interaction in between $h_{2}s$ molecules are the permanent dipole-dipole interactions that develop due to the dipole that exists in between the polar h - s bond. the positively charged hydrogen atom is attracted towards the negative end of the dipole that is sulfur of the adjacent $h_{2}s$ molecule. therefore, these forces of attraction are relatively stronger than that of london dispersion forces that exist in non polar molecules.

Answer from: Quest

Yes you are correct use the scientific method

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Use the thermodynamic data at 298 k below to determine the ksp for barium carbonate, baco3 at this t...

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