The equilibrium NH3(aq) + H20(1) -NH4*(aq) + OH (aq) at 25 *C is subjected to a temperature jump which slightly increases the concentration of NH4*(aq) and OH (aq). The measured relaxation time is 7.61 ns. The equilibrium constant for the system is 1.78 x 10-5 at 25 *C, and the equilibrium concentration of NH3(aq) is 0.15 mol dm-3. (c) Calculate the rate constant for the reverse reaction. Krev =. Just value in 3 sig. fig., in normal or exponential format, e.g. 1.16E6 meaning 1.16 x 106 must use capital E. Choose a unit in the next question. . QUESTION 10 (d) choose a unit for the reverse rate constant. O no unit O 5-1 O Umolis O L2mol?is

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The equilibrium NH3(aq) + H2O() –NH4*(aq) + OH"(aq) at 25 °C is subjected to a temperature jump
which slightly increases the concentration of NH4* (aq) and OH (aq).
The measured relaxation time is 7.61 ns.
The equilibrium constant for the system is 1.78 × 10-5 at 25 *C, and the equilibrium concentration of
NH3(aq) is 0.15 mol dm-3.
(c) Calculate the rate constant for the reverse reaction.
Krev =.
meaning 1.16 x 106 must use capital E. Choose a unit in the next question. .
Just value in 3 sig. fig., in normal or exponential format, e.g. 1.16E6
QUESTION 10
(d) choose a unit for the reverse rate constant.
O no unit
s-1
L'mol/s
L2mor?is
I know the answer is 4.0x10^10 L/mol/s but
do not know how to get the answer.
Transcribed Image Text:Chemistry The equilibrium NH3(aq) + H2O() –NH4*(aq) + OH"(aq) at 25 °C is subjected to a temperature jump which slightly increases the concentration of NH4* (aq) and OH (aq). The measured relaxation time is 7.61 ns. The equilibrium constant for the system is 1.78 × 10-5 at 25 *C, and the equilibrium concentration of NH3(aq) is 0.15 mol dm-3. (c) Calculate the rate constant for the reverse reaction. Krev =. meaning 1.16 x 106 must use capital E. Choose a unit in the next question. . Just value in 3 sig. fig., in normal or exponential format, e.g. 1.16E6 QUESTION 10 (d) choose a unit for the reverse rate constant. O no unit s-1 L'mol/s L2mor?is I know the answer is 4.0x10^10 L/mol/s but do not know how to get the answer.
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