Part A) Part A) Consider the following reaction: 2NO(g)+2H2(g)N2(g)+2H2O(g). Wha

ID: 933207 • Letter: P

Question

Part A)

Consider the following reaction:
2NO(g)+2H2(g)N2(g)+2H2O(g).

What is the reaction rate at 1000 K when the concentration of NO is increased to 0.14 M, while the concentration of H2 is 1.10×102M?

Express your answer using two significant figures

Part B)

If a solution containing 34.0 g of a substance reacts by first-order kinetics, how many grams remain after three half-lives?

Express your answer with the appropriate units.

Part C)

The reaction 2NO22NO+O2 has the rate constant k= 0.63 M1s1.

If the initial concentration of NO2 is 0.112 M, how would you determine how long it would take for the concentration to decrease to 3.30×102M?

Express your answer using two significant figures.

Part: B: Giiven the initial amount of the reactant, A0 = 34.0 g

For a chemical reaction, half-life is defined as the time after which the initial amount of the reactant becomes half.

Hene after 1st half-life the amount of the reactant remains = 34.0 g / 2 = 17.0 g

After 2nd half-life the amount of the reactant remains = 17.0 g / 2 = 8.50 g

After 3rd half-life the amount of the reactant remains = 8.50 g / 2 = 4.25 g (answer)

Hence the amount of the substance remained after 3 half-lives = 4.25 g

Part:C: The given reaction is

------------------------2NO2 ------ > 2NO+O2

Initial conc(M): 0.112 M ------ 0 ----- 0

Conc.after 't' sec:(0.112 - y)M, y M, y M

Hence initial concentration of NO2, a = 0.112 M

Concentration after time 't', (a - y) = 3.30x 10-2 M

=> y = 0.112 M - 3.30x 10-2 M = 0.079 M

rate constant, k = 0.63 M-1s-1

Since the unit of k is M-1s-1, this is an example of second order reaction.

Hence for second order reaction,

Rate = d[NO2] / dt = - k x [NO2]2

The integrated rate law for second orer reaction is

t = (1 / k) x [y / a x (a - y)]

=> t = (1 / 0.63 M-1s-1) x [0.079 M / 0.112 M x 3.30x 10-2 M] = 34 seconds (answer)

Part-A: This much information is not sufficient enough to answer this question