Can anyone help me to solve this
When 10cm3 of 1 mol dm^-3 nitric acid is mixed with 20cm3 of 1 mol dm^-3 sodium hydroxide solution, there is a temperature rise of (delta)T.If the reaction is repeated with 20cm3 of nitric acid of 1 mol dm^-3 and 20cm3 of 1 mol dm^-3 sodium hydroxide, the temperature rise is A 2(delta)TB 1.5(delta)TC (delta)T D 0.75(delta)TThe answer is B
HNO3 + NaOH NaNO3 + H2O
In the first reaction, n HNO3 = 0.01 mol and n NaOH = 0.02 mol... Hence n H2O = 0.01 mol as HNO3 is the limiting reagent in this reaction
However, when the reaction is repeated, n HNO3 is 0.02 mol, hence n H2O = 0.02 mol
So, the amount of energy released will increase by a factor of two
Also, the mass of the system has increased
The first reaction involved 30 cm3 of solution, and the second reaction involved 40 cm3 of solution
So, the mass has increased by a factor of
Since Q=mcΔT, ΔT =
c is constant, hence hasn't increased/decreased between reactions, so can be omitted ΔT =
As you can see, ΔT is proportional to Q and inversely proportional to m... So, ΔT will increase by a factor of two, but decrease by a factor of , giving an overall increase of 1.5
Furthermore, you can use the above equation to verify this, by substituting the factors in: ΔT = = 1.5
Hence, regardless of your approach, you'll see that ΔT has increased by a factor of 1.5
In the first reaction, n HNO3 = 0.01 mol and n NaOH = 0.02 mol... Hence n H2O = 0.01 mol as HNO3 is the limiting reagent in this reaction
However, when the reaction is repeated, n HNO3 is 0.02 mol, hence n H2O = 0.02 mol
So, the amount of energy released will increase by a factor of two
Also, the mass of the system has increased
The first reaction involved 30 cm3 of solution, and the second reaction involved 40 cm3 of solution
So, the mass has increased by a factor of
Since Q=mcΔT, ΔT =
c is constant, hence hasn't increased/decreased between reactions, so can be omitted ΔT =
As you can see, ΔT is proportional to Q and inversely proportional to m... So, ΔT will increase by a factor of two, but decrease by a factor of , giving an overall increase of 1.5
Furthermore, you can use the above equation to verify this, by substituting the factors in: ΔT = = 1.5
Hence, regardless of your approach, you'll see that ΔT has increased by a factor of 1.5
(edited 4 years ago)
Thank you ...
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