A catalyst:
a. actually participates in the reaction.
b. changes the equilibrium concentration of the products.
c. does not affect a reaction energy path.
d. always decreases the rate for a reaction.
e. always increases the activation energy for a reaction. http://www.chem.tamu.edu/class/fyp/mcquest/ch16.html
Re: CHEM LAB C
Posted: December 21st, 2014, 6:44 pm
by alwaysmatts
bernard wrote:A catalyst:
a. actually participates in the reaction.
b. changes the equilibrium concentration of the products.
c. does not affect a reaction energy path.
d. always decreases the rate for a reaction.
e. always increases the activation energy for a reaction. http://www.chem.tamu.edu/class/fyp/mcquest/ch16.html
is it
a. even though it does not change during the reaction
Re: CHEM LAB C
Posted: February 22nd, 2015, 11:57 pm
by bernard
alwaysmatts wrote:
bernard wrote:A catalyst:
a. actually participates in the reaction.
b. changes the equilibrium concentration of the products.
c. does not affect a reaction energy path.
d. always decreases the rate for a reaction.
e. always increases the activation energy for a reaction. http://www.chem.tamu.edu/class/fyp/mcquest/ch16.html
is it
a. even though it does not change during the reaction
Correct, your turn! Sorry I took a while.
Re: Chemistry Lab C
Posted: June 9th, 2015, 6:02 pm
by bernard
A question for next year's topic, Gas Laws!
Solve for the gas constant in L atm/mol K, and solve the following question with your result. A sample of argon gas at STP occupies 56.2 liters. Determine the number of moles of argon and the mass in the sample. Show your work.
Re: Chemistry Lab C
Posted: June 9th, 2015, 9:54 pm
by jkang
bernard wrote:A question for next year's topic, Gas Laws!
Solve for the gas constant in L atm/mol K, and solve the following question with your result. A sample of argon gas at STP occupies 56.2 liters. Determine the number of moles of argon and the mass in the sample. Show your work.
Part I: The gas constant (R) is defined as 8.314 J/molK. Converting that to Latm/molK, you just need to convert from J to Latm, which has a conversion factor of 0.009869. Thus, R = 0.08206 Latm/molK.
Part II: The ideal gas law is PV = nRT. STP gives a temperature of 273.15 K and pressure of 100 kPa (.9869 atm) under IUPAC. Applying this equation to the problem, you get (.9869 atm)(56.2 L) = n(0.08205 Latm/molK)(273.15 K), and solving for n gives the value of 2.47 moles (3 sig figs due to volume). Multiplying by the molar mass of argon, this gives a total of 98.9 grams.
Re: Chemistry Lab C
Posted: June 9th, 2015, 10:12 pm
by bernard
Correct! Here's another solution to the first part, which was what I had in mind.
1 mol of a gas at STP is 22.4 L. Plugging in this and the values for STP (273.15 K and 1 atm) into the ideal gas law, PV = nRT: (1 atm)(22.4 L) = (1 mol)(273.15 K)R. R = (1 atm)(22.4 L)/(1 mol)(273.15 K) = 0.082006 L atm/mol K.
Re: Chemistry Lab C
Posted: July 23rd, 2015, 8:17 pm
by RandomPerson
Oooh, gas laws. I can actually contribute to this.
I dunno how much SciOly will get into gas laws above PV = nRT, the next step up is Van Der Waals. So here's a question that always showed up on my Gen Chem Tests:
Is the Van Der Waals equation of state. You'll notice there are two extra terms, one with the Pressure and one with the Volume. These attempt to adjust for the non ideality of real gases (though... even VDW isn't that great in reality).
So a couple precursor questions:
1. What does the assumption of an ideal gas mean? (How do we assume they interact?)
2. For what values of the three state variables (P,V,T) is the ideal gas equation a reasonable approximation? (low or high for all three).
Finally the payoff:
3. Each of the parameters a, and b account for something. What do the parameters "mean"? What do they try to adjust for/what do they account for?
You should be able to answer this just by analyzing the EOS.
That's a bit vague so here's an example format. Pretend there is an EOS: PV = nR(T + c*n). I could say that c accounts for the extra heat one mol of this real gas has compared to an ideal gas. This is not a valid EOS, of course.