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Katarina [22]
3 years ago
5

Consider the potassium permanganate reaction again.

Chemistry
2 answers:
Rus_ich [418]3 years ago
6 0
Hey there !

Mole ratio :

<span>2 KMnO4 + 16 HCl → 2 KCl + 2 MnCl2 + 8 H2O + 5 Cl2

2 moles KMnO4 ----------------- 8 moles H2O
3.45 moles KMnO4 ------------- (moles H2O )

Moles H2O = 3.45 * 8 / 2

Moles H2O = 27.6 / 2

 = 13.8 moles of H2O

</span>The option that was given is wrong , <span>You're right.</span>
puteri [66]3 years ago
3 0

Answer:

Moles of H2O produced = 13.8

Explanation:

Given:

Moles of KMnO4 reacted = 3.45

To determine:

moles of H2O produced

Explanation:

Given reaction:

2KMnO4 + 16HCl →2KCl + 2MnCl2 + 8H2O + 5Cl2

Based on the reaction stoichiometry:

2 moles of KMnO4 produces 8 moles of H2O

Therefore, moles of H2O produced when 3.45 moles of KMnO4 react is:

= \frac{3.45\ moles\ KMnO4 * 8\ moles\ H2O}{2\ moles\ KMnO4} = 13.8

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CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7
padilas [110]

<u>Answer:</u> The \Delta G for the reaction is 54.425 kJ/mol

<u>Explanation:</u>

For the given balanced chemical equation:

CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)

We are given:

\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol

To calculate \Delta G^o_{rxn} for the reaction, we use the equation:

\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]

For the given equation:

\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]

Putting values in above equation, we get:

\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J

Conversion factor used = 1 kJ = 1000 J

The expression of K_p for the given reaction:

K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}

We are given:

p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm

Putting values in above equation, we get:

K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85

To calculate the gibbs free energy of the reaction, we use the equation:

\Delta G=\Delta G^o+RT\ln K_p

where,

\Delta G = Gibbs' free energy of the reaction = ?

\Delta G^o = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = 8.314J/K mol

T = Temperature = 25^oC=[25+273]K=298K

K_p = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol

Hence, the \Delta G for the reaction is 54.425 kJ/mol

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A. Hydrogen Bonding

Explanation:

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