3. What is the standard cell potential for the spontaneous reaction between thes

Question

3. What is the standard cell potential for the spontaneous reaction between these two half cells? 3e Al 1.66 V Ag' e-Ag E 0.80 V 24. The series of figures shows the reaction 2A Proceeding to dynamic Product equilibrium which figure also represents equilibrium for this reaction at the same temperature? All boxes represent the same o O 25. For an exothermic reaction performed at two different temperatures, which condition gives the higher concentration of products at equilibrium? (Assume that is the only variable.) (A) the reaction at the higher temperature (B) the reaction at the lower temperature (cy the concentrations will be the same because K is a (D) not enough information to answer below. The equilibrium concentrations at 250 C were found to be: Col 0.0960 M.[Hal .0.191 M and ICH,0Hj 0.150 M. What is K for the reaction? (A) 428 (C) 0.122 D) 0.0233 affect the equilibrium constant expression? The value of K, is (A) X. (B) (C) 2X. (D) 4x 28. What is the equilibrium constant expression for this reaction? (A) 29. A flask initially contains 0.100 nitrogen and 0.240 M oxygen. The equilibrium constant, Ke for the reaction is 4.08x10 at this temperature. What is the equilibrium concentration of NO? (A) 1.6x10 M (B) 3.1x10 M (C) 0.34 M 0.68 M 30. Which chemical equation has K. K, 31. Which shifts to the left side (reactants) upon a decrease in volume? (A) 30.0g 20,0g)

Explanation / Answer

23) For Al Standard reduction potential is –1.66 V which is –ve and hence Al will act as a reductant and undergo oxidation i.e. Al act as anode and hence Ag as cathode.

Then,

E0cell = E0cathode – E0anode

E0cell = 0.80 – (– 1.66)

E0cell = 2.46 V

Answer- option-(A).

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24) From the figures given its clear that

At equilibrium [product] = 2 x [reactant]

The same is the situation in option figure (A)

Figure in option (A) represent equilibrium condition at same temperature.

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25)

Exothermic reaction: In an exothermic reaction heat is evolved and according to Le Chatelier’s principle at lower temperature product formation will be favored. Because at lower temperature more and more heat will be absorbed by surrounding and to minimize this lowering of heat more product formation will occur.

Hence.

Answer- (B) the reaction at lower temperature will have higher product concentration.

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26) Formation of H3COH (Methanol)

CO (g) + 2H2 (g) <-----------> H3COH (g)

Expression for equilibrium constant Kc for this reaction is,

Kc = [H3COH]/[H2]2[CO]

Where [H3COH], [H2] and [CO] are equilibrium concentrations.

Let us put there given values in above expression for Kc.

Kc = (0.150) / [0.0960 x (0.191)2]

Kc = 42.8

Answer – Option-(A)

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27)

Given reaction: CH4 (g) + 2H2O (g) < ----------> CO2 (g) + 4H2 (g)

Expression for equilibrium constant is,

Kp = [CO2][H2]4/[H2O]2[CH4]

But Kp = X ………… (given)

So, [CO2][H2]4/[H2O]2[CH4] = X ……………….(1)

For the reaction : 2CH4 (g) + 4H2O (g) < ----------> 2CO2 (g) + 8H2 (g)

Expression for equilibrium constant is,

Kp’ = [CO2]2[H2]8/[H2O]4[CH4]2

Kp’ = ([CO2][H2]4/[H2O]2[CH4])2

Kp’ = (Kp)2

Kp’ = X2   …………..(from eq. (1))

Answer – Option (B)

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28) In Kc expression concentrations of species from all phases are included, hence,

Kc = [CO2]6[H2O]6/ [O2]6[C6H12O6]

Answer – option-(D)

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29)

Given chemical reaction:

N2 (g) + O2 (g) < ----------> 2NO (g)

Kc = [NO]2/[N2][O2] …………..(1)

Let at equilibrium ‘2X’ M of product NO formed. Then by stoichiometry we write,

[NO] = ‘2X’ M, [N2] = (0.100 – X ) M, [O2] = (0.240 – X) M.

Kc = 4.08 x 10–4

Let us put all these values in eq. (1)

4.08 x 10–4 = (2X)2 / [(0.100 – X) (0.240 – X)

X <<< 0.1 and X <<<< 0.240 hence small concentration assumption holds true.

So,

4.08 x 10–4 = 4X2 / (0.100 x 0.240 )

4.08 x 10–4 = 4X2 / 0.0240

X2 = 4.08 x 10–4 x 0.0240/4

X2 = 2.448 x 10–6

X = 1.56 x 10–3 M

Now,

[NO] = 2 x X = 2 x 1.56 x 10–3

[NO] = 3.1 x 10–3 M

Answer- Option-(B)

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30)

Kp = Kc x (RT)n

When n = 0

We have Kp = Kc

n = Number of gaseous product – Number of gaseous reactants.

With n = 0

0 = Number of gaseous product – Number of gaseous reactants

Number of gaseous product = Number of gaseous reactants

I.e. for the reaction in which this is true, Kp = kc holds.

For H2 (g) + I2 (g) <--------> 2HI (g)

We have n = 0

Hence for this reaction Kp = Kc.

Answer –option-(D)

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31)

According the Le Chatelier’s principle,

If volume of a system in equilibrium is lowered equilibrium shift in a direction that will decrease the number of moles of gaseous species.

A decomposition reaction,

2KClO3 (g) <--------> 2KCl (s) + 3O2 (g)

Backward reaction i.e. reactant side is accompanied with lowering of number of gaseous species and on lowering volume KClO3 i.e. reactatnt formation will be favored.

Answer – Option –(C).

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