Question

A + 2 B → C + 2 D
is known to be first order in [A] and second
order in [B]. When 0.20 moles of A and 0.30
moles of B are placed in a 2.00 liter container,
it is observed that 7 × 10−5 moles of C are
formed per second with no change in volume.
What is the rate constant at 25◦C?
Answer in units of L2/mol2· s

Answer 1

Answer:

Write the rate of reaction in terms of the rate of disappearance of reactant and the rate of formation of products:

\(NO_{(g)} + O_{3 (g)} \rightarrow NO_{2(g)} + O_{2(g)}\)

\(2C_2H_{6 (g)} + 7O_{2(g)} \rightarrow 4 CO_{2(g)} + 6 H_2O_{(aq)}\)

\(H_{2 (g)} + I_{2 (g)} \rightarrow 2HI_{(g)} \)

\(4OH_{(g)} + H_2S_{(g)} \rightarrow SO_{2(g)} + 2H_2O_{(aq)} + H_{2(g)}\)

S9.1

\(\text{rate of reaction} = \dfrac{-∆[NO]}{∆t} = \dfrac{-∆[O_3]}{∆t} = \dfrac{∆[NO_2]}{∆t} = \dfrac{∆[O_2]}{∆t} \)

\(\text{rate of reaction} = \dfrac{-∆[C_2H_6]}{2∆t} = \dfrac{-∆[O_2]}{ 7∆t} =\dfrac{∆[CO_2]}{4∆t} = \dfrac{∆[H_2O]}{6∆t} \)

\(\text{rate of reaction} = \dfrac{-∆[H_2]}{ ∆t} = \dfrac{-∆[I_2]}{∆t} = \dfrac{∆[HI]}{2∆t} \)

\(\text{rate of reaction} = \dfrac{-∆[OH]}{4∆t} = \dfrac{-∆[H_2S]}{∆t} = \dfrac{∆[SO_2]}{∆t} = \dfrac{∆[H_2O]}{∆t} = \dfrac{∆[H_2]}{∆t} \)

9.2: Reaction Order

Q9.2

Determine the value of the rate constant for the elementary reaction:

\[I_{2(g)} + H_{2 (g)} \rightarrow 2HI_{(aq)}\]

given that when [Br2] is 0.15 M and [H2] is 0.2M, the rate of reaction is 0.005 M s-1 at 298 K.

S9.2

rate of reaction = k[Br2][H2]

0.005 Ms-1 =k ( 0.15M)2(0.2M)

k= 1.11 M-1s-1

Q9.3

Given the rate of the third order reaction:

\[A + B + C \rightarrow P \]

is 0.05 Ms-1

If the [A] = 0.05 M, [B] = 0.01M, and [C] = 0.25M. What is the third order rate constant?

S9.3

rate of reaction = k[A][B][C]

0.05Ms-1 =k (0.05M)(0,01M)(0.25M)

k= 0.05Ms-1/( 0.05) (0.01)( 0.25) M3

k= 400M-2s-1