Print Crk ulator Periodic Table Question 17 of 22 Map to Sapling Learning The fo

Question

Print Crk ulator Periodic Table Question 17 of 22 Map to Sapling Learning The following reaction is a single-step, bimolecular reaction: CH,Br+NaOH --> CH3OH + Na Br When the concentrations of CH3Br and NaOH are both 0.175 M, the rate of the reaction is 0.0050 Mis. (a) What is the rate of the reaction if the concentration of CHaBr is doubled? Number M/s (b) What is the rate of the reaction if the concentration of NaOH is halved? Number M/s (c) What is the rate of the reaction if the concentrations of CHjBr and NaOH are both increased by a factor of three? Number M/s Hint Previous ® Gw-up & View Sklilion e> Check Answer D Next Exit

Explanation / Answer

a) rate of the reaction is given by

rate = k [CH3Br][NaOH]

Where, k is rate constant

k = 0.0050(M/s)/(0.175M×0.175M)

= 0.1633M-1s-1

When concentration of CH3Br doubled

rate = k ×(2 ×[CH3Br])×[NaOH]

rate = 0.1633M-1s-1×(2×0.175M)×(0.175M)

= 0.0100M/s

b) when concentration of NaOH halved

rate = k × [ CH3Br] × (1/2×[NaOH])

= 0.1633M-1s-1 × 0.175M ×(1/2×0.175M)

= 0.0025M/s

c) when concentration of both CH3Br and NaOH increased by a factor of 3

rate = k ×(3×[CH3Br])×(3×[NaOH])

= 0.1633M-1s-1 ×3 ×0.175M ×3 ×0.175M

= 0.0450M/s

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.