Question

9. A flask contains XeFx(g) at 25 torr. Hydrogen gas is added to the flask until the total pressure in the flask is 10 0 torr. Following the (spark catalyzed) reaction, the flask contains only xenon gas, hydrogen fluoride gas and unreacted hydrogen gas. Following removal of the HF (g) (by reaction with solid sodium hydroxide), the flask contains only xenon and hydrogen at a total pressure of 50 torr. Name the compound, XeFx Assume constant temperature 

10. Nitric oxide (NO) reacts with molecular oxygen as follows: 2 NO (g)+ O2 (g) → 2 NO2 (g) 

Initially NO and O2 are separated as shown below. When the valve is opened, the reaction quickly goes to completion. Determine what gases remain at the end and calculate their partial pressures. Assume that the temperature remains constant temperature. 

Answer 1

9)

2 XeFx + x H2 => 2 Xe + 2x HF

From reaction stoichiometry:

Moles of H2 reacted/moles of XeFx reacted = x/2


Ideal gas equation: PV = nRT

At constant T and V => moles n are proportional to P


P(XeFx) reacted = P(XeFx) initial = 25 torr


From reaction stoichiometry, moles of Xe formed = moles of XeFx reacted

=> P(Xe) final = P(XeFx) reacted = 25 torr


P(H2) final = (P(Xe) final + P(H2) final) - P(Xe) final

= 50 - 25 = 25 torr


P(H2) initial = P(total) - P(XeFx) initial

= 100 - 25 = 75 torr


P(H2) reacted = P(H2) initial - P(H2) final

= 75 - 25 = 50 torr


Moles of H2 reacted/moles of XeFx reacted = P(H2) reacted/P(XeFx) reacted

= 50/25 = 2


Thus x/2 = 2 => x = 4

The compound is XeF4 (xenon tetrafluoride)

10)

FOR NO

PV = nRT

n = (PV)/(RT) = (0.500atm *4.00L)/(0.0821* 298K) = 0.0817moles

For O₂:

PV = nRT

n = (PV)/(RT) = (1.00atm * 2.00L)/(0.0821 * 298K) = 0.0817moles

From stoichiometry, the number of moles of NO₂formed is 0.0817 moles.

Looking at the balanced equation, the limiting reagent is NO.After the reaction, NO will no more. Therefore, the partial pressure of NO is0atm.

The ratio of moles is related to the ratio of partialpressures. Since only half of the moles of O₂ arereacted, half of it is left over (that is, 0.0817/2 = 0.04085moles).

Now we use PV=nRT again to find the total pressure:

PV = nRT

P = (nRT)/V = ((0.04085 moles + 0.0817 moles) *0.0821*298K)/(4.0L + 2L) = 0.4997 atm

Now we can use ratios:

x + 2x = 0.4997

3x = 0.4997

x = 0.167

Therefore:

Partial PressureO₂= 0.167 atm

Partial PressureNO₂ = 0.333 atm