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Ksivusya [100]
3 years ago
5

Which of the following reactions are redox reactions? 4Li(s)+O2(g)→2Li2O(s) Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s) Sr(NO3)2(aq)+Na2SO4(aq

)→ SrSO4(s)+2NaNO3(aq) HBr(aq)+KOH(aq)→H2O(l)+KBr(aq)
Chemistry
1 answer:
Lapatulllka [165]3 years ago
4 0

Answer:  

As2O3 → H3AsO4

As2O3 → 2H3AsO4             balance As

5H2O + As2O3 → 2H3AsO4       balance O by adding H2O to one side

5H2O + As2O3 → 2H3AsO4 + 4H+    balance H by adding H+ to one side

5H2O + As2O3 → 2H3AsO4 + 4 H+ + 4e-     balance charge by adding electrons to one side

 

Now do the same for the other part of the reaction

NO3- → NO

NO3- → NO + 2H2O

4H+ + NO3- → NO + 2H2O

3e- + 4H+ + NO3- → NO + 2H2O

 

Now cancel the electrons by multiplying the first equation by 3 and the second equation by 4, then add them together .

 

3As2O3 + 4NO3- + 7H2O + 4H+ → 6H3AsO4 + 4NO

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Give the direction of the reaction, if K >> 1. Give the direction of the reaction, if K >> 1. The forward reaction i
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A. for K>>1 you can say that the reaction is nearly irreversible so the forward direction is favored. (Products formation)

B. When the temperature rises the equilibrium is going to change but to know how is going to change you have to take into account the kind of reaction. For endothermic reactions (the reverse reaction is favored) and for exothermic reactions (the forward reaction is favored)

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A. The equilibrium constant K is defined as

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In any case  

aA +Bb  equilibrium Cd +dD

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K= \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}}

[] is molar concentration.

If K>>> 1 it means that the molar concentration of products is a lot bigger that the molar concentration of reagents, so the forward reaction is favored.

B. The relation between K and temperature is given by the Van't Hoff equation

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K_{2}=\frac{K_{1}}{e^{\frac{-deltaH^{o}}{R}*(\frac{1}{T_{1}} -\frac{1}{T_{2}})}}

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For exothermic reactions when we increase the temperature the denominator in the equation would have a negative exponent so K_{2} is greater that K_{1} and the forward reaction is favored.

When we have an endothermic reaction we will have a positive exponent so K_{2} will be less than K_{1} the forward reactions is not favored.  

{e^{\frac{-deltaH^{o}}{R}*(\frac{1}{T_{1}} -\frac{1}{T_{2}})}}

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