CH 14 pt 2 The reaction AB( a q )A( g )+B( g ) is second order in AB and has a r

Question

CH 14 pt 2

The reaction AB(aq)A(g)+B(g) is second order in AB and has a rate constant of 0.0143 M1s1 at 25.0 C. A reaction vessel initially contains 250.0 mL of 0.128 M AB which is allowed to react to form the gaseous product. The product is collected over water at 25.0 C.

Part A

How much time is required to produce 195.0 mL of the products at a barometric pressure of 743.3 mmHg . (The vapor pressure of water at this temperature is 23.8 mmHg.)

Dinitrogen pentoxide decomposes in the gas phase to form nitrogen dioxide and oxygen gas. The reaction is first order in dinitrogen pentoxide and has a half-life of 2.81 h at 25 C.

If a 1.6 L reaction vessel initially contains 750 torr of N2O5 at 25 C, what partial pressure of O2will be present in the vessel after 200 minutes?

Explanation / Answer

Answer of Part -A

First of all , we assume both A and B are gases

The we calculate P(A+B).. (Pressure A + Pressure B)

P(A+B) = Ptotal - PH2O = 743.3mmHg - 23.8mmHg = 719.5 mmHg

PA = 1/2 P(A+B) = 359.75mmHg
nA = PA x V/(RT) = 359.75mmHg x (1atm/760mmHg) x 0.195L / (0.08206 Latm/moleK x 298.15K)
nA = 0.00377 moles A

Now, from the balanced equation, 1 mole AB ---> 1 mole A
so. moles AB reacted = 0.00377 moles

Now initial moles AB = 0.2500L x (0.128 moles / L) = 0.03200 moles
remaining moles AB = 0.03200 - 0.00377 = 0.02823
final molarity = 0.02823 moles / 0.2500L = 0.1130M

and from here...assuming second order...
- d[A] / dt = k x [A]²
1 / [A]² d[A] = -k dt
1 / [A]² d[A] = -k dt...
-1/[At] + 1/[Ao] = -kt
1/[At] = +kt + 1/[Ao]

so..
t = (1/[At] - 1/[Ao]) / k
t = (1/0.1130M - 1/0.128M) / (0.0143 / Msec) = 72.52 sec

t = 72.52 sec

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