For the reaction below, if the rate of appearance of Br_2 is 0.180 M/s, what is

Question

For the reaction below, if the rate of appearance of Br_2 is 0.180 M/s, what is the rate of disappearance of NOBr? NOBr(g) rightarrow 2 NO(g) + Br_2(g) 0.360 M/s 0.090 M/s 0.180 M/s 0.720 M/s 2.0 M/s What is the name given to the point of highest potential energy that is reached during the course of a chemical reaction? catalyst reactant intermediate activated complex product A reaction A + B rightarrow C + D has DeltaH = -80 KJ and has E_act =15 KJ. What is E_act for the reaction C + D rightarrow A + B ? 15 KJ 95 KJ 65 KJ -15 KJ 30 KJ For which of the following rate laws would the units of the rate constant have the general form L^2 middot mol^-2 middot sec^-1 ? rate = k [A]^3 rate = k[A]^4 rate = k[A]^2 rate = k[A] rate = k

Explanation / Answer

11) Given balanced stoichiometric chemical reaction:

2NOBr (g) -----------> 2 NO (g) + Br2 (g)

From the stoichometry we can write,

d[Br2]/dt = ½ d[NO]/dt = – ½ d[NOBr]/dt

Hence,

– ½ d[NOBr]/dt = d[Br2]/dt

i.e.

½ rate of disappearence of NOBr = rate of appearance of Br2 ………..(word disappearance takes –ve sign)

So,

Rate of disappearencee of NOBr = 2 x rate of appearance of Br2

Rate of disappearance of NOBr = 2 x 0.180 M/s

Rate of disappearance of NOBr = 0.360 M/s

Option-A

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12) The point with highest potential energy is that reached during course of the reaction is named as Activated complex.

Intermediate is less higher point of potential energy.

Option-D)

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13) Given reaction: A + B ------> C + D

With H = – 80 kJ and Eact = 15 kJ

–ve sign of enthalpy indicate that heat is evolved out from reactant A and B to reach to the products C and D.

For the reverse reaction C + D -----------> A + B

We first have to supply heat energy to prepare molecule to react which is exactly + 80 kJ and then activated comlex formation require more +15 kJ to reach to highest Potential energy point.

Hence totally, 80 + 15 = 95 kJ of energy needed to be supplied.

Option-B

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14)

A) rate = k [A]3

k = rate / [A]3

k = rate x [A]–3

Rate is in M/L.s

[A] is in M/L

Hence,

k = M/L.s x (M/L)–3

k = M x L–1 x s–1 x M–3 x L3

k = M–2 x L2 x s–1

i.e. unit of k = L2 . mole–2. s–1

Is the required unit and hence

Option-A)

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