enthalpy ( year 12 ) Flashcards
what about delta T?
CHANGE IN TEMPERATURE IS SAME FOR KELVIN AND CELCSIUS !
why is it hard to determine standard enthalpy change of formation of hexane directly ?
many different hydrocarbons would form OR activation energy too high OR reaction too slow OR they don’t react together
standard state of cylcloalaknaes ( eg hexane )
LIQUID!
Why would the actual conditions used in the chemical industry be differetn from the question ? (2)
IF 2 MARKS PRESENT, THEN STATE TWO DIFFERENT THINGS !
high pressure as fewer moles (of gas) on right-hand side
low temperature as (forward) reaction is exothermic
remember: REVERSIBLE REACTIONS DONT REALLY HAVE REACTANTS AND PRODUCTS (JUST SYA RIGHT HAND / LEFT HAND SIDE )
actual conditions differ because it may be too expensive to use high pressure (1) and reaction would go very slowly if low temp was used (1)
STATE THE PROPOGATION REACTIONS
Draw the enthalpy profile diagram for a catalysed exothermic reaction and an uncatalysed one ?
NOTE: ADD ARROWS TO Ea (catalysed and uncatalysed) OR YOU WONT GET MARK
why are catalysts sustainable ( 4)
. they arent used up in reaction
.catalysts find an alternative reaction route with a lower activation energy (1) therefore lower temperatures are used (1) thus less carbon dioxide is emitted into the atmosophere (1)
extra makrs
catalysts / enzymes generate specific products !
calculate the enthalpy change of reaction
AND WHAT SIGN IS IT ?
Enthalpy change of reaction = - 135 ( NEVER FORGET SIG FIGS )
WHY IS IT MINUS ? because in the question it says energy is released and temp of solution icnreases thus the reaciton isd exothermic hterfore negative
what happens when pressure is increased here? (2)
more concentrated (particles)= more collisions per second !
( I put no change in reaction rate due to equal moels on both sides but thats apparently wrong ! )
what does boltzmann distribution consist of ? (3)
x axis- energy
y axis- number of particles
Activation energy does not change (1)
Fewer molecules have energy above activation energy (1)
So fewer successful collisions (1)
NOTE: for enthalpy of reaction, multiply (delta H)= (q/moles) BY THE AMOUNT OF MOLES IN THE BALANCED EQUATION!
EQ: 2NH4SCN(s) + Ba(OH)2(s) Ba(SCN)2(s) + 2H2O(l) + 2NH3(g)
question; calculate enthalpy change of reaction
DO NOT FORGET THE POSITVIE SIGN
Define averagew bond enthalpy
energy required to break one mole of specified type of bond in a gaseous molecule !
explain bonding in double bond
overlap of 2 P orbitals to form a Pi bond !
Suggest a reason why the actual value for the enthalpy change of this reaction is different from the calculated value.
Bond enthalpies may not be the same as the average bond enthalpy OR The idea that bonds have different strengths in different environments
