mean/average energy needed to break 1 mole of bonds in gaseous state
so CH4(g) -> C(g) + 4H(g)
the enthalpy change here would be around 1663 kJ mol-1
but since 4 bonds of C-H are broken then u divide by 4 to get the average
so the mean bond enthalpy here would be 1663/4
correct me if im wrong!
so CH4(g) -> C(g) + 4H(g)
the enthalpy change here would be around 1663 kJ mol-1
but since 4 bonds of C-H are broken then u divide by 4 to get the average
so the mean bond enthalpy here would be 1663/4
correct me if im wrong!
1) non standard conditions
2) incomplete combustion
3) heat lost too surrowndings
2) incomplete combustion
3) heat lost too surrowndings
Original post by laura273
give reasons why the standard enthalpy change of combustion of methanol will be different from that previously calculated
Original post by RickHendricks
mean/average energy needed to break 1 mole of bonds in gaseous state
so CH4(g) -> C(g) + 4H(g)
the enthalpy change here would be around 1663 kJ mol-1
but since 4 bonds of C-H are broken then u divide by 4 to get the average
so the mean bond enthalpy here would be 1663/4
correct me if im wrong!
so CH4(g) -> C(g) + 4H(g)
the enthalpy change here would be around 1663 kJ mol-1
but since 4 bonds of C-H are broken then u divide by 4 to get the average
so the mean bond enthalpy here would be 1663/4
correct me if im wrong!
Yes that’s correct. It’s important to note the use of the word “average” as each C-H bond will require different amounts of energy to remove as each hydrogen will be in a different environment when the other is removed. Hence the word mean or average.
Peak of the curve = most probable energy
Slightly to right of peak = mean energy.
Number of particles (area under graph) is always the same.
Increase in temperature = more particles have higher kinetic energies = more particles likely to have activation energy or higher in a given instant = increase frequency of successful collisions = rate of reaction increases.
Slightly to right of peak = mean energy.
Number of particles (area under graph) is always the same.
Increase in temperature = more particles have higher kinetic energies = more particles likely to have activation energy or higher in a given instant = increase frequency of successful collisions = rate of reaction increases.
Original post by laura273
the rate of a chemical reaction increases as the temperature is increased because there are more collisions. discuss to what extent this statement is true and describe any other factors that should be considered when looking at the effect of temperature on rate of reaction
... and your question is?
Original post by laura273
explain the effect of a catalyst on the rate of a reaction using a Boltzmann distribution graph which shows the energies of molecules in a gas at a fixed temperature
Would it not be better to do your own homework?
You might actually learn something ...
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