Nitrogen monoxide reacts with oxygen to give nitrogen dioxide: 2 NO(g) + O 2(g)

Question

Nitrogen monoxide reacts with oxygen to give nitrogen dioxide:

2 NO(g)  + O2(g) -----> 2 NO2(g)

The rate law is -delta[NO]/delta[t] = k[NO]2[O2], where the rate constant is 1.16*10-5 M-2*s-1 at 339oC. the initial pressure oh NO and O2 are 155 mmHg and 345 mmHg, respectively. What is the rate of decrease of partial pressure of NO (in mmHg per sec)?

(Hint: from the ideal gas law, obtain an expresion for the molar consentration (mol/L) of a particular gas in terms of its partial pressure.)

Nitrogen monoxide reacts will oxygen to give nitrogen dioxide 2 NO_g + O_out rightarrw 2 NO_ The rate law is Delta(NO)/Deltat = k(NO)^2(Q_2). Where the rate containt is 1.16 times 10^-5 M^3 d^-3 at 335degreeC. The initial pressures of NO and Q_2 are 455 mmh/g Hg 345 mmh/g, respectively. What is the ratio of decreate of pressure of NO [in mmHg per section)?

Explanation / Answer

The rate law is given as, -delta[NO]/delta[t] = k[NO]2[O2]

and k is given as 1.16 x 10-5 M-2 s-1 and T = 339 + 273 = 612 K

From ideal gas equation, PV = nRT

If we divide both sides by V, then

P = (n/V)RT

where (n/V) is concentration in mol/L. So, the equation could also be written as...

P = CRT

if both sides are divided by RT, then..

C = P/RT

for NO gas, 155 mm Hg x 1atm/760 mm Hg = 0.204 atm

concentration, C for NO gas = 0.204/(0.0821 x 612) = 0.00406 M

likewise, for O2 gas, = 345 mm Hg x 1atm/760 mm Hg = 0.454 atm

Concentration of O2 gas = 0.454/(0.0821 x 612) = 0.00904 M

Plug in the values in the given rate law...

-delta[NO]/delta[t] = k[NO]2[O2] = 1.16 x 10-5 M-2 s-1 (0.00406M)2 (0.00906M) = 1.73 x 10-12 M s-1

This could again be converted in terms of pressure.

P = CRT

let's write molarity (M) as mol L-1

1.73 x 10-12 mol L-1 s-1 x 0.0821 atm L mol-1 K-1 x 612 K = 8.69 x 10-11 atm s-1

Convert atm into mm Hg

8.69 x 10-11 atm s-1 x 760 mm Hg/1atm = 6.60 x 10-8 mm Hg s-1 (Answer)

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