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2 years ago

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Consider the balanced chemical equation below.

1 answer:

viktelen [127]2 years ago

5 0

Explanation:

$here \: is \: your \: explanation : - \\ \\ given \: balanced \:equation \: = > \\ \\ 2 A=>C \: + \: 4D \\ \\ by \: this \: equation \: we \: get \: \\ \\ 2 \: moles \: of \: A \: produce \: 4 \: moles \: \\ \\ of \: D \\ \\ hence \: . \: 1 \: mole \: can \: produce \: = 4 \div 2 \\ \\ = > 2 \: moles \: \\ \\ so \: if \: 6 \: moles \: of \: A \: used \: then \: \\ \\ amount \: of \: D \: produced \: = (6 \times 2) \\ \\ = > 12 \: moles \: of \: D \\ \\ \mathcal\blue{ Hope \: it \: helps \: you \: (. ❛ ᴗ ❛.) }$

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Explanation:

Let the volume of the solution be 100 ml.

As the volume of glycol = 50 = volume of water

Hence, the number of moles of glycol = $\frac{mass}{molar mass}$

= $\frac{density \times volume}{molar mass}$

= $\frac{1.1088 \times 50}{62 g/mol}$

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Hence, number of moles of water = $\frac{50 \times 0.998}{18}$

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As glycol is dissolved in water.

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= 17.9

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2 years ago

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I2(g) + Cl2(g)2ICl(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.62 mol

galina1969 [7]

Answer:

The change in entropy of the surrounding is -146.11 J/K.

Explanation:

Enthalpy of formation of iodine gas = $\Delta H_f_{(I_2)}=62.438 kJ/mol$

Enthalpy of formation of chlorine gas = $\Delta H_f_{(Cl_2)}=0 kJ/mol$

Enthalpy of formation of ICl gas = $\Delta H_f_{(ICl)}=17.78 kJ/mol$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]$

For the given chemical reaction:

$I_2(g)+Cl_2(g)\rightarrow 2ICl(g),\Delta H_{rxn}=?$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(2\times \Delta H_f_{(ICl)})]-[(1\times \Delta H_f_{(I_2)})+(1\times \Delta H_f_{(Cl_2)})]$

$=[2\times 17.78 kJ/mol]-[1\times 0 kJ/mol+1\times 62.436 kJ/mol]=-26.878 kJ/mol$

Enthaply change when 1.62 moles of iodine gas recast:

$\Delta H= \Delta H_{rxn}\times 1.62 mol=(-26.878 kJ/mol)\times 1.62 mol=-43.542 kJ$

Entropy of the surrounding = $\Delta S^o_{surr}=\frac{\Delta H}{T}$

$=\frac{-43.542 kJ}{298 K}=\frac{-43,542 J}{298 K}=-146.11 J/K$

1 kJ = 1000 J

The change in entropy of the surrounding is -146.11 J/K.

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3 years ago