View Full Version : Please help me with this Chemistry question. Please.
kraftykitten
Jan 27, 2011, 10:15 AM
At a pressure of 1 atmosphere and a temperature of 25 °C, 500 cm3 of the gas HX has a mass of 0.400 g. Calculate the value of the relative atomic mass of X and hence identify element X. (You should assume that 1 mole of gas occupies 24.5 dm3, and 1 dm3 = 1000 cm3. Assume the relative atomic mass of hydrogen to be 1.0.)
(ii)
The enthalpy change for Reaction 1 is −1180 kJ mol−1. Sketch a fully labelled diagram showing how the internal energy changes for this reaction. On your diagram indicate the energy barrier for the reaction.
DrBob1
Jan 27, 2011, 04:16 PM
Meow. You have the information needed to determine the molecular weight of HX and hence of X. (You know the volume of one mole of gas, and you know how much it weighs) Show us your work and we will help you where you are having problems.
kraftykitten
Jan 28, 2011, 04:28 AM
At a pressure of 1 atmosphere and a temperature of 25 °C, 500 cm3 of the gas HX has a mass of 0.400 g. Calculate the value of the relative atomic mass of X and hence identify element X. (You should assume that 1 mole of gas occupies 24.5 dm3, and 1 dm3 = 1000 cm3. Assume the relative atomic mass of hydrogen to be 1.0.)
(ii)
The enthalpy change for Reaction 1 is âˆ'1180 kJ molâˆ'1. Sketch a fully labelled diagram showing how the internal energy changes for this reaction. On your diagram indicate the energy barrier for the reaction.
Relative atomic mass
NH (g) + 2X (g) = N (g) + 4HX (g) (Equation is balanced)
500cm3 1000cm3
2X no. of mols = volume x concentration
= 1000 x 10-3 x 1
= 1 mol dm-3
Concentration NH = moles/volume
= 1 mol x 10-3 dm-3/500 x 10-3 dm-3
= 0.002 mol dm-3
No. of mol NH = volume x concentration
= 24.5 dm3 x 0.002 mol dm-3
= 4.81 mol
= 0.5 x 4.81
= 2.41
After this, I became completely confused. I've tried looking through my notes & going over them, but I do feel completely lost in this question. I am not sure if my methods are correct or if I am totally going about it the wrong way. It took me four hours yesterday to do this part, but my mind is drawing a blank because I am confused with this. Please could you help me Dr. Bob or anyone please. Thank you. I would really appreciate it.
Unknown008
Jan 28, 2011, 07:38 AM
At a pressure of 1 atmosphere and a temperature of 25 °C, 500 cm3 of the gas HX has a mass of 0.400 g. Calculate the value of the relative atomic mass of X and hence identify element X. (You should assume that 1 mole of gas occupies 24.5 dm3, and 1 dm3 = 1000 cm3. Assume the relative atomic mass of hydrogen to be 1.0.)
You are told that 1 mol of gas occupies 24.5 dm^3
You have 500 cm^3 of gas.
24.5 dm^3 -> 1 mol
1 dm^3 -> 1/24.5 mol
1000 cm^3 -> 1/24.5 mol
500 cm^3 -> (1/24.5)/2 mol
Now, this number of moles has a mass of 0.4 g.
Find the mass of 1 mole.
1 mole of H has a mass of 1 g. Hence, find the atomic mass of X.
The enthalpy change for Reaction 1 is -1180 kJ mol^-1. Sketch a fully labelled diagram showing how the internal energy changes for this reaction. On your diagram indicate the energy barrier for the reaction.
Is it possible for us to know what is the 'reaction 1' the question is referring to?
Also, the equation you gave contains strange compounds...
NH (g) + 2X (g) = N (g) + 4HX (g) (Equation is balanced)
NH doesn't exist so far I know... if X is alone, it means surely it's a noble gas? Then N is alone? That is very difficult to find because it will violently react with another N atom in the surrounding to form N2.
And lastly, the equation, as strange as it is, is not balanced.
1 N on the left, 1 N on the right;
1 H on the left, 4 H on the right;
2 X on the left, 4 X on the right.
=/
kraftykitten
Jan 28, 2011, 09:07 AM
Please please help me with this as I have been trying for days now to do this but can't understand how to it. Even if someone could help me with how I should go about working it out would be a great help so that I would understand it. I need help before Tuesday 1st February please. Guidelines would be very much appreciated. I didn't put in the correct equation before, but I have done so below:
Gaseous hydrazine reacts with a gas X to form the gases nitrogen and HX by way of the following reaction: (nb: HX is the only information given)
N^2H^4 (g) + 2X^2 (g) = N^2 (g) + 4HX (g) (the ^ indicates the number is below line)
(I)
At a pressure of 1 atmosphere and a temperature of 25 °C, 500 cm3 of the gas HX has a mass of 0.400 g. Calculate the value of the relative atomic mass of X and hence identify element X. (You should assume that 1 mole of gas occupies 24.5 dm3, and 1 dm3 = 1000 cm3. Assume the relative atomic mass of hydrogen to be 1.0.)
(ii)
The enthalpy change for Reaction 1 is −1180 kJ mol−1. Sketch a fully labelled diagram showing how the internal energy changes for this reaction. On your diagram indicate the energy barrier for the reaction.
(When part (ii) is talking about Reaction 1 - this is all the information it gives in this question).
Please do help or guide me if you can as I am working through other questions for chemistry. But I cannot do this and my methods were wrong. I will help others in the future if I can.
Unknown008
Jan 28, 2011, 09:30 AM
Please, do not start another thread when you already have the same question going on. Reply on this thread itself by typing your reply in the 'Answer Box'. This avoids confusion and is easier for yourself as everything is in one single thread, that is the help is not scattered around the forum. Thank you.
I already give you help for part (i). Could you show what you have using what I told you?
Also, it would be good if you typed in manually the symbols as they don't get entered properly if you copy paste them.
Ok, the equation now is:
N_2H_4\ (g) + 2X_2\ (g) \rightarrow N_2\ (g) + 4HX\ (g)
Is the enthalpy \Delta H = -1180\ kJmol^{-1} ?
You need to draw an energy level diagram with N2H4 + X2 on one horizontal line of a certain energy, then N2 + 4HX on another lower horizontal line. Then, you need to draw a curve going from the first line, upwards until a maximum then downwards to the other line, a little like this:
http://www.avogadro.co.uk/h_and_s/hprofile.gif