Question

2. Using the data in the table, determine the rate constant of the reaction and select the appropriate units.

A+2B⟶C+D

Trial	[A] (M)[A] (M)	[B] (M)[B] (M)	Rate (M/s)
1	0.340	0.340	0.0127
2	0.340	0.680	0.0127
3	0.680	0.340	0.0508

k=

Units

3. The rate constant for this first‑order reaction is 0.480 s−1 at 400 ∘C.

A⟶products

How long, in seconds, would it take for the concentration of A to decrease from 0.650 M to 0.270 M?

4. After 49.0 min, 31.0% of a compound has decomposed. What is the half‑life of this reaction assuming first‑order kinetics?

t1/2= min

Answer 1

-"enlivo- 1 let oder of a reaction is 22 0042 of reacon with desbeelfonis - overal reaction order - n +2B uty CAD -Rall(r) LEAT" [87 pom tasek TATTOO I Talal 1 tedal, [A= 10.397 - B1 =10.34] 00127 MS poto 2 000/20K 0.34 I tal-2 a G./27 K [0.3472x0947² am go and w We obserbed order01 sachon Bis zerolo). And. Trials * 4x012.2[0:050 3 ) = (2x034 1 10.5970 from er 1 and TV Twe obersed oroler of reaction 'A' is two (2) Henee Order of reaction is n= urty - 270-2 puting the value of M&y an eno on Qahy - Khois722 0.3470 "K = 0.0127 415 0.109 49 0.3 40 oran

*enlivo Griven, lel rate Constan L isk K = 0.4805 intratial concentration : 0.65M at Ernest -con contrakoon 0.27M A products at tao 0.650 o at tit o.270 We kosow shat 2.303 log t Sio tarul It is a expression for 1st order reactor a = initial concenfation Tu & cencontralien at time to SK = 2:303 lo J. (0:56 to_2203 Logos -log??? ¢ loyment t 2303 [1.81-1987 Colyes = 1.89 (= 2303 X0.88 107 27 = 1.98 f-120 second

-"enlivo Date: Pago 47 biren, 7 = $9 mot decomposed – 31.7. We know that K> 2803. 209 for a 1160-31869 K = 2.303 loy 100 E 10 69 R 2, 2013 103.40 – 10769] K = 2.30 3 X2X1 - 1.837 ( 10g69 =183/ K= 2303 0.17 1611021) R = 0.0079 int are Ye now El for 1st order reaction is (4 = 0.693 49 to c 0.693 0.0079 87:72 minuti