For the reaction N20g) 3 H20gj 2 NH3(g), the reaction rate in terms of 1. the ch

Question

For the reaction N20g) 3 H20gj 2 NH3(g), the reaction rate in terms of 1. the change in concentration of H20g) would be D) 1/3 A[H2]/At B) -1/3 A[H2/At C) ADH2]/At A) -ADH2]/At 2. A 1.0 x 103 g/mL water solution of human hemoglobin(Hb) (MM 64,000) has the same osmotic pressure as a 2.5 x 10-3 g/mL water solution of protein X. The MM of protein X is AD 64,000 B 25,600 C) 128,000 D) 160,000 E) cannot be determined 3. Referring to the above solutions, the intermolecular forces in protein X, compared to the forces in Hb, are A) greater than Hb B) the same as Hb C) less than Hb D) cannot be determined The endothermic reaction A (g) B(g) C(g) is at equilibrium. Predict the effect of each of the following changes in the equilibrium [B] by choosing the correct letter from the following B) decreases [B] A) increases Bl C) no effect on [B] 4. Add A 5. Remove some C 6. Raise the temperature 7. Decrease the volume of the container 8. Add a catalyst 9. The conversion of A to B is a first order reaction with ti/2 -10 minutes. Starting with 0.300 MA, the time required for the [A] to reach 0.075 M is A) 10 min B) 20 min C) 30 min D) impossible to tell

Explanation / Answer

1.

N2(g) + 3H2(g) ------------> 2NH3(g)

The reaction in terms of H2 will be given as

r = - 1/3 * delta[H2]/delta(t) because stoichiometric coefficient of H2 is 3 which we have to divide and since it is a reactant so its concentration will decrease because of which the - ve sign is provided

2.

Osmotic pressure = c * R * T

So,

c1 * R * T = c2 * R * T    ( Here 1 represents Hb and 2 represents X )

So,

c1 = c2

So,

(mass /Volume*MM)1 = ( mass / Volume*MM)2

(mass/Volume)1 * 1/MM1 = (mass / Volume)2 * 1/MM2

1 x 10^-3 g/mL * 1/64000 = 2.5 x10^-3 * 1/MM2

MM2 = 2.5 * 64000 = 160,000

Answer is (D) 160,000

3.

(A) Intermolecular forces in protein X is greater than that of Hb because molar mass of protein X is greater so the intermolecular forces will be stronger

A(g) <-------------> B(g) + C(g)

Kc = [B] [C] / [A]

4.(A)

If we add [A] then concentration of [B] will increase because more [A] will react to form more [B] in order to keep Kc constant.

5.(B)

If we remove some C then [B] will have to decrease to keep Kc constant

6.(A)

Since reaction is endothermic so increase in temperature will increase the formation of products so [B] will increase

7.(B)

delta(ng) = moles of gaseous products - moles of gaseous reactants = 2 - 1 = 1

So a decrease in volume of container will shift the reaction to the left because delta(ng) is positive (+1) . So [B] will decrease

8.(C)

Adding a catalyst has no effect in concentration of substances

9.

A --------> B    

For first order reaction,

t(1/2) = 0.693 / k     and   ln(At / A0) = - k * t

Now,

t(1/2) = 10 min

k = 0.693 / 10 = 0.0693 min^-1

Now,

ln(At / A0) = - k * t

ln( 0.075 / 0.300) = - 0.0693 * t

t = 20 min

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