Originally posted by hoay:The products obtained by cracking an alkane, X, are methane, ethene and propene.
The mole fraction of ethene in the products is 0.5.
What is the identity of X?
A C6H14 B C8H18 C C9H20 D C11H24
UltimaOnline replied:
Start by drawing these 3 smaller hydrocarbon products and piece them together like a jigsaw puzzle. Based on the fact that the longest hydrocarbon obtained is only 3 carbons, obviously the original large alkane must be branched. Based on the fact that 50% of the products are ethene, that means two of the four branches must be ethyl groups. The third branch must be the propyl group. The intersection of the branches, itself generates methane.
Based on this help I have come up with this structure:
CH2-CH3
CH3-C-CH2-CH3
CH2-CH3
Is it possible to calcultae the remaining mole fraction of mentahne and propene from the data given. Since they both will be 0.5- (x+y) = 1.
Mole fraction refers to the relative moles of a particular species, over the total number of moles of all species. Hence, the mole fractions are
50% or 1/2 are ethene,
25% or 1/4 are methane,
and 25% or 1/4 are propene.
Sir Thank u very much for a quick reply.
What factors can affect the value of the activation energy of a reaction?
1 the presence of a catalyst
2 changes in temperature
Catalyst lowers the activation energy of the reaction.
Temperature raises the average K.E of the molecules hence they achieve Ea. So both these factors are correct. Am I right??
Originally posted by hoay:What factors can affect the value of the activation energy of a reaction?
1 the presence of a catalyst
2 changes in temperatureCatalyst lowers the activation energy of the reaction.
Temperature raises the average K.E of the molecules hence they achieve Ea. So both these factors are correct. Am I right??
No, activation energy for any particular reaction is only altered by the use of a catalyst, and remains unchanged by temperature.
Only when activation energy (Ea) remains unchanged, then the % of reactant molecules with energy equals or exceeding Ea can increase as temperature increases, thus increasing the rate of reaction.
An analogy would be : if your income goes up, but your cost of living also goes up, then the rate at which you can reach your retirement goals won't be able to increase.
Only when the cost of living (ie. activation energy) remains unchanged, and your income (ie. % of molecules with energy equals or exceeding Ea) goes up, only then will the rate at which you can reach your retirement goals happily increase.
X is a salt of one of the halogens chlorine, bromine, iodine, or astatine (proton number 85).
The reaction scheme shows a series of reactions using a solution of X as the starting reagent.
X HNO3/ AgNO3 a precipitate excess NH3(aq) colorless solution
excess HNO3
precipiate
What could X be?
A sodium chloride
B sodium bromide
C potassium iodide
D potassium astatide
It can be wither Nacl or NaBr uptill the colorless solution but when is happeneing on the addition of excess HNO3.??
The answer is A.
Originally posted by hoay:X is a salt of one of the halogens chlorine, bromine, iodine, or astatine (proton number 85).
The reaction scheme shows a series of reactions using a solution of X as the starting reagent.X HNO3/ AgNO3 a precipitate excess NH3(aq) colorless solution
excess HNO3
precipiate
What could X be?
A sodium chloride
B sodium bromide
C potassium iodide
D potassium astatideIt can be wither Nacl or NaBr uptill the colorless solution but when is happeneing on the addition of excess HNO3.??
The answer is A.
HNO3 simply removes the NH3 ligands by protonating them, regenerating the silver halide ppt.
The key is the "excess NH3(aq)" used, instead of "excess concentrated NH3(aq)" that is required to dissolve AgBr(s). Hence it can be deduced that the ppt is AgCl(s), rather than AgBr(s).
The Haber process for the manufacture of ammonia is represented by the following equation.
N2(g) + 3H2(g) 2NH3(g) ΔH = –92 kJ mol–1
Which statement is correct about this reaction when the temperature is increased?
A Both forward and backward rates increase.
B The backward rate only increases.
C The forward rate only increases.
D There is no effect on the backward or forward rate.
When the temperature is increased heat is absorbed by the product NH3 since forward reaction is exothermic and it starts converting to the reactants so backward rate increases. so B is is the correct answer. But A is the naswer.
The Haber process for the manufacture of ammonia is represented by the following equation.
N2(g) + 3H2(g) <---> 2NH3(g) ΔH = –92 kJ mol–1
Which statement is correct about this reaction when the temperature is increased?
A) Both forward and backward rates increase.
B) The backward rate only increases.
C) The forward rate only increases.
D) There is no effect on the backward or forward rate.
Answer : A
When the temperature is increased, the percentage of reactant molecules with energy equal or exceeding the activation energies for both the backward and forward reactions, are increased (along with temperature). Hence, both the forward and backward reaction rates increase.
However, the percentage increase in rate is greater for the reaction in the direction with the larger activation energy. If the forward reaction is endothermic (ie. the forward reaction will accordingly have a larger activation energy, compared to the backward reaction's smaller activation energy), then increasing the temperature will increase both the forward and backward reaction rates, but will increase the forward reaction rate by a larger magnitude compared to the backward reaction rate. This results in the position of equilibrium shifting towards the right (ie. towards the products), even as the rate of both the forward and backward reactions simultaneously increase.
In an endothermic forward reaction, heat may be regarded as a reactant (rather than product). Increasing the temperature is thus akin to increasing the concentration of the reactants, which will therefore cause the position of equilibrium to shift towards the right (ie. towards the products), as predicted by Le Chatelier's principle (ie. the system attempting to remove the added heat energy by using some of it to convert the reactants to the products). Mathematically, the Kc (ie. equilibrium constant) value will increase, while the Qc (ie. reaction quotient) value remains initially unchanged.
Hence (while increasing the temperature of the system increases both the forward and backward rate constants), the rate of the forward reaction (ie. indicated by the new relatively much larger k-forward value) will now exceed the rate of the backward reaction (ie. indicated by the new relatively only slightly larger k-backward value) until equilibrium is eventually re-established for the system, ie. Qc = Kc.
Once Qc = Kc and the system has re-established equilibrium, the rates of the forward and backward reactions will be exactly the same. This is despite the k-forward rate constant still having a larger value compared to the k-backward rate constant. To understand this, recall that the rate of reaction is given as the rate constant multiplied by the molarities of the reactants, each raised to the power of their orders. Thus at position of equilibrium, having smaller reactant molarities and a larger k-forward rate constant, will be mathematically equal to having larger product molarities and a smaller k-backward rate constant.
Excellent explanation.
I did not consider the word "rate"- I was thinking about the position of equilibrium of the equilibrium in favour of reactants. Your explanation should be a part of my notes in Haber's process.
1- Another confusion is regarding the yield of NH3 under the expalnation u have discussed above which will not increase neither decrease.
2-If we increase the pressure in Haber's process more ammonia would be formed it means that the value of Kp should increase because the [NH3] is increasing so the value should go up. Secondly, if we look into the use of stanadard electrode potential values to this pressure-increase case then in E-nought case [reactants] increae the [product] so the value incraeses. I am mixing the two concepts beacuse i am not getting the idea that increasing pressure will not increase the Kp of a reaction.
Very much thanks.
The heat Iiberated in the neutralisation given below is -114kJ/mol.
2NaOH(aq) + H2SO4(aq) Na2SO+(aq) + 2HzO(l)..........1
By using this information, what is the most likely value for the heat liberated in the following
neutralisation?
Ba(OH)2(aq) + 2HCl(aq) BaCl2(aq) + 2H2O(l)...............2
A -57kJmol-l B -76kJmol-1 C -114kJmol-1 D -228kJmol
(ans)
Eq 1 are both strong acid and strong baseas well as in eq 2 both are strong acid and relatively strong base so the energy would be same Is this the right conclusion??
Originally posted by hoay:The heat Iiberated in the neutralisation given below is -114kJ/mol.
2NaOH(aq) + H2SO4(aq) Na2SO+(aq) + 2HzO(l)..........1
By using this information, what is the most likely value for the heat liberated in the following
neutralisation?Ba(OH)2(aq) + 2HCl(aq) BaCl2(aq) + 2H2O(l)...............2
A -57kJmol-l B -76kJmol-1 C -114kJmol-1 D -228kJmol
(ans)
Eq 1 are both strong acid and strong baseas well as in eq 2 both are strong acid and relatively strong base so the energy would be same Is this the right conclusion??
Yes, all acids and bases used in both experiments are all strong, and the same number of moles of H2O are formed. This is potentially an issue of ambiguity, because while enthalpy of neutralization technically should be per mole of H2O formed anyway, regardless of how many moles of water were formed. But sometimes, the question setter and mark scheme may expect different enthalpy values for different moles of water generated.
Cambridge for instance, committed a similar ambiguity in the Singapore 2010 H2 Chem paper 1, when the student was tasked with calculating the entropy change for the reaction. There are two possible answers (one for per mole of the species, the other for per molar quantities of the equation), both arguably correct, but Cambridge only accepted one of the two possible answers. To this day, many school teachers disagree with Cambridge's choice of the correct answer.
1- In the reaction:
SO2 + 2H2S 2H2O + 3S
how do we know that Sulfur is acting as a reducing agent or an acid since the oxidation number of S has changed to zero. (from S+4 and S -2)
2- Which statement about relative atomic mass is correct.
A It is a ratio of masses.
B It is measured in grams
C It is related to the number of atoms in a a molecule
D It is the same as the mass as of 1 mol of atoms.
A is correct since the average atomic mass of an element is compared with 12 g of C12 so it ia ratio with respect to C12. Is this right?
Originally posted by hoay:1- In the reaction:
SO2 + 2H2S 2H2O + 3S
how do we know that Sulfur is acting as a reducing agent or an acid since the oxidation number of S has changed to zero. (from S+4 and S -2)
2- Which statement about relative atomic mass is correct.
A It is a ratio of masses.
B It is measured in grams
C It is related to the number of atoms in a a molecule
D It is the same as the mass as of 1 mol of atoms.
A is correct since the average atomic mass of an element is compared with 12 g of C12 so it ia ratio with respect to C12. Is this right?
Q1.
To be precise, sulfur dioxide is being reduced (and is hence the oxidizing agent) and hydrogen sulfide is being oxidized (and is hence the reducing agent). Since elemental sulfur is not present as a reactant, it is less proper to say "sulfur is a reducing agent" or "sulfur is an oxidizing agent".
This reaction may be regarded as the opposite of a proper disproportionation reaction, because of the two sulfur-containing reactant species, one is being oxidized and the other being reduced, both generating elemental sulfur as a product. Thus it would be more proper to say, "sulfur in SO2 is being reduced, while sulfur in H2S is being oxidized, both generating elemental sulfur as a product".
.
Bronsted-Lowry acid-base reaction
H2S functions as the Bronsted-Lowry acid and donates protons. The O atoms of SO2 function as the Bronsted-Lowry base and accepts protons.
.
Lewis acid-base reaction
The S atoms of H2S function as the Lewis base (ie. nucleophile). The S atoms of SO2 function as the Lewis acid (ie. electrophile).
.
.
Q2.
Yes, you're totally correct.
Many crude oils contain sulphur as H2S. During refining, by the Claus process, the H2S is converted into solid sulphur, which is then removed.
reaction I 2H2S(g) + 3O2(g) → 2H2O(l) + 2SO2(g)
reaction II 2H2S(g) + SO2(g) → 2H2O(l) + 3S(s)
Which statements about the Claus process are correct?
1 H2S is oxidised in reaction I.
2 SO2 oxidises H2S in reaction II.
3 SO2 behaves as a catalyst.
1 is correct and 3 is incorrect. So we are left with 2. Can we say that H2S in reaction II has lost protons and gained Oxygen so it is being oxidized. Is it correct??
Originally posted by hoay:Many crude oils contain sulphur as H2S. During refining, by the Claus process, the H2S is converted into solid sulphur, which is then removed.
reaction I 2H2S(g) + 3O2(g) → 2H2O(l) + 2SO2(g)
reaction II 2H2S(g) + SO2(g) → 2H2O(l) + 3S(s)Which statements about the Claus process are correct?
1 H2S is oxidised in reaction I.
2 SO2 oxidises H2S in reaction II.
3 SO2 behaves as a catalyst.1 is correct and 3 is incorrect. So we are left with 2. Can we say that H2S in reaction II has lost protons and gained Oxygen so it is being oxidized. Is it correct??
It is best to use Oxidation States (also known as Oxidation Numbers) when determining what's been oxidized and what's been reduced. Statement 2 is correct because SO2 is being reduced (OS change of S from +4 to 0) and H2S is being oxidized (OS change of S from -2 to 0).