Chapter 14 Home works correct for the chemical equatiom given concerning relativ

Question

Chapter 14 Home works correct for the chemical equatiom given concerning relative ofreaction CHONG is equal to the rate of disapperar The rate of disappearance of of appearance of of CHOH or Hao is four times the Eate of di 2- which relationship, corect, compares the rates of the following reactants and products? 3. Ozone decomposes to oxygen according to the balanced chemical equation below. If the 2050g) 3 formation of oxygen? rate of disappearance ofozone is -2.4 x 10 Ms. what is the rate of a. -2.4 x 10 MWs b. 1.6 x 06 Ms c. 2.4 10 Ms 3.6 x 10 M/s c. 7.2 x 10 M/s 4. For the reaction below, if the rate of appearance of Brz is 0.180 M/s, what is the rate of disappearance of NOBr? 2 NOBr(g) 2 NO(g) Br(g) a. -0.360 MWs b. -0.090 Ms 0.090 M/s 0.180 M/s c. 0.720 MW's

Explanation / Answer

(1)

2 CH3OH (g) + 3 O2 (g) --------> 2 CO2 (g) + 4 H2O (g)

Rate = -(1/2)d[CH3OH]/dt = - (1/3)d[O2]/dt = + (1/2)d[CO2]/dt = +(1/4)d[H2O]/dt

So, the correct option is,

(d) The rate of appearance of H2O is two times the rate of appearance of CO2

(2) 2 NOCl (g) --------> 2 NO (g) + Cl2 (g)

(c) rate = - (1/2)d[NOCl]/dt = + (1/2)d[NO]/dt = d[Cl2]/dt

(3) 2O3 (g) ---------> 3 O2 (g)

Rate = - (1/2) d[O3]/dt = +(1/3)d[O2]/dt

(1/2)(2.4*10-6) = (1/3)d[O2]/dt

d[O2]/dt = 3.6 * 10-6 M/s

(4) 2 NOBr (g) ---------> 2 NO (g) + Br2 (g)

Rate = -(1/2)d[NOBr]/dt = +(1/2)d[NO]/dt = d[Br2]/dt

-(1/2)d[NOBr]/dt = 0.180

d[NOBr]/dt = - 0.36 M/s

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