Nitrogen and hydrogen, with volumes of 12.0 L each, are injected into a sealed rigid reaction chamber containing a catalyst. The pressure of the chamber is 8.00 atmospheres and at a known temperature. The reaction is allowed to proceed to completion while being maintained at a constant temperature. How many liters of each gas is present after the reaction occurs?
What is the partial pressure of each gas after the reaction goes to completion?
Which of the following statements about an ideal gas is incorrect?
A. There is no interaction between the particles of gas.
B. The gas particles occupy negligible volume.
C. The "free volume" is the same as the volume of the container.
D. Gases behave ideally under conditions of standard temperature and pressure.
E. Gases under conditions of low pressure and high temperature are closest to "ideal."
I need these answered ASAP (w/i 24 hrs)!!!!
Please answer quickly, even if you only know one of the question's answers.
Thanks in advance!! :)
5 answers
You need a Keq (Kc or Kp) for the reaction.
I would look at D as being the untrue statement.
The problem in 1 says that it proceeds to "completion" which means to me that we don't worry about the Kc or Kp.
N2 + 3H2 ==> 2NH3. Since both reactants are given we know it is a limiting reagent problem. This is a gas problem; therefore, we are allowed to use L directly as moles.
L NH3 formed if N2 is limiting = 12*2 = 24L.
L NH3 formed if H2 is limiting = 12*(2 moles NH3/3 moles H2) = 12(2/3 = 8 L.
The smaller value is always the correct one to chose; therefore, we will have 8 L NH3 formed and 12 L H2 used and H2 is the limiting reagent.
How much H2 used? All of it; therefore, there is zero H2 remaining.
How much N2 remains? How much is used.
N2 used 8L NH3 x (1 mole N2/2 moles NH3) = 4 moles N2 used. Subtract from initial to find that remaining.
N2 + 3H2 ==> 2NH3. Since both reactants are given we know it is a limiting reagent problem. This is a gas problem; therefore, we are allowed to use L directly as moles.
L NH3 formed if N2 is limiting = 12*2 = 24L.
L NH3 formed if H2 is limiting = 12*(2 moles NH3/3 moles H2) = 12(2/3 = 8 L.
The smaller value is always the correct one to chose; therefore, we will have 8 L NH3 formed and 12 L H2 used and H2 is the limiting reagent.
How much H2 used? All of it; therefore, there is zero H2 remaining.
How much N2 remains? How much is used.
N2 used 8L NH3 x (1 mole N2/2 moles NH3) = 4 moles N2 used. Subtract from initial to find that remaining.
Thanks for question 1. So what would the partial pressure be for the gases?
I got 2.67 and 5.33 (b/c I did 3x = 8 atm (2x for NH3 and x for N2). Is that on the right track?
1 answer