Actual Rate Law from a Table of Concentration/Rate Data 27....

Actual Rate Law from a Table of Concentration/Rate Data 27. Given the following data, determine the rate law for the reaction $NH_4^+(aq) + NO_2^-(aq) \rightarrow N_2(g) + 2H_2O(l)$ Experiment $[NH_4^+]$ (M) $[NO_2^-]$ (M) Rate (M/s) 1 0.2500 0.2500 $1.25 \times 10^{-4}$ 2 0.5000 0.2500 $2.50 \times 10^{-4}$ 3 0.2500 0.1250 $6.25 \times 10^{-5}$ a. $k[NH_4^+][NO_2^-]$ d. $k[NH_4^+]^0[NO_2^-]^1$ b. $k[NH_4^+][NO_2^-]^1$ e. $k[NH_4^+][NO_2^-]^2$ c. $k[NH_4^+]^1[NO_2^-]^{1/2}$

Transcript

00:01 Hello hi students let me explain to the question here we have to write the rate law for the reaction so the reaction here it is s2 08 2 minus acquiesce plus 2 iodine minus aqueous it gives rise to 2 s .o 4 2 minus aqueous plus iodine acquies so here what is meant by rate law rate law or rate equation so it is an equation it is an equation it links it links the forward reaction rate forward reaction rate with concentrations or pressures of reactants with concentrations or pressures of reactants of reactants and constant parameters and constant parameters so here whenever we see the so here rate law.

01:10 So the rate law or rate it is given by the formula that is k into concentration of a and concentration of b to the power of n.

01:20 So here whenever we see the experiment number 2 here is 2 .o8 2 minus ion concentration and i minus ion concentration in experiment 2 it is 0 .010 here 0 .0 .0 .0 .0.

01:39 In experiment 4, the concentrations are 0 .020, here 0 .18.

01:47 So, here, whenever we see the ratio, rate of ratio of these reactants, it is k into 0 .0 .012 whole to the power of m, 0 .09 hold to the power of n.

02:02 K.

02:02 0 .020 whole to the power of m, 0 .18 whole to the power of n, that is rate of equation 2 by rate of equation 4, that is equal to 0 .5 to the power of m, 0 .5 to the power of m and that is equal to rate 2 by rate 4.

02:27 So here, so here rate is equal to rate equal to the change in iodine concentration by d t, that is of s2o32 minus by t.

02:45 Because here, so here whenever we see the equation, the moles of the, the moles of s2o8 is one, but iodine is two.

02:57 So the ratio here it is 1 is to 2.

03:00 So that's why here, the iodine, it is equal to half s2 -o -3 -2 minus byte t is equal to, that is equal to rate 2.

03:18 Rate 2 is equal to half into 0 .10 divided by, that is equal to 1 .19 into 10 to minus 4, m...

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