Why do you divide by 0.025 x 2 and not 0.05 x2?

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heptanitrocubane
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I am confused as to why, once calculating the energy change in kJ (-2.863) you then divide by (0.025 x 2) and not (0.05 x 2) as that is the volume of the solution times the molarity

25 cm3 2M nitric acid and 25cm3 2M potassium hydroxide. A temp rise of 20.2 to 33.9, calculate deltaH given specific heat capacity of water is 4.18.

I get to -2.863 / (0.05 x 2) = -28.6 instead of dividing by (0.025 x 2) to get -57.3 which is the answer.
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Davies Chemistry
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(Original post by heptanitrocubane)
I am confused as to why, once calculating the energy change in kJ (-2.863) you then divide by (0.025 x 2) and not (0.05 x 2) as that is the volume of the solution times the molarity

25 cm3 2M nitric acid and 25cm3 2M potassium hydroxide. A temp rise of 20.2 to 33.9, calculate deltaH given specific heat capacity of water is 4.18.

I get to -2.863 / (0.05 x 2) = -28.6 instead of dividing by (0.025 x 2) to get -57.3 which is the answer.
The equation is KOH + KNO3 --> KNO3 + HNO3

The heat released in the beaker is -2.863 KJ as you say

Delta H is the heat released "per mole of equation" i.e when one mole of KOH reacts with one mole of HNO3 (not the combined moles of KOH and HNO3)

so delta H= -2.863/0.05
= -57.3 KJ

Think of another example:
Na2CO3 + 2HCl --> Na2CO3 + CO2 + H2O
Imagine you react 0.1 mol Na2CO3 with 0.2 mol HCl,..
Then DeltaH = Q/0.1

You wouldn't divide by 0.2 moles (or 0.2 + 0.1) ..Delta H is the heat released when 1 mole of Na2CO3 reacts with 2 moles of HCl...i.e. "per mole of equation"

It is a common point of confusion
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charco
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(Original post by Davies Chemistry)
The equation is KOH + KNO3 --> KNO3 + HNO3

The heat released in the beaker is -2.863 KJ as you say

Delta H is the heat released "per mole of equation" i.e when one mole of KOH reacts with one mole of HNO3 (not the combined moles of KOH and HNO3)

so delta H= -2.863/0.05
= -57.3 KJ

Think of another example:
Na2CO3 + 2HCl --> Na2CO3 + CO2 + H2O
Imagine you react 0.1 mol Na2CO3 with 0.2 mol HCl,..
Then DeltaH = Q/0.1

You wouldn't divide by 0.2 moles (or 0.2 + 0.1) ..Delta H is the heat released when 1 mole of Na2CO3 reacts with 2 moles of HCl...i.e. "per mole of equation"

It is a common point of confusion
You might want to correct that
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Pigster
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(Original post by charco)
You might want to correct that
Maybe the acid is in excess?

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heptanitrocubane
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(Original post by Davies Chemistry)
The equation is KOH + KNO3 --> KNO3 + HNO3

The heat released in the beaker is -2.863 KJ as you say

Delta H is the heat released "per mole of equation" i.e when one mole of KOH reacts with one mole of HNO3 (not the combined moles of KOH and HNO3)

so delta H= -2.863/0.05
= -57.3 KJ

Think of another example:
Na2CO3 + 2HCl --> Na2CO3 + CO2 + H2O
Imagine you react 0.1 mol Na2CO3 with 0.2 mol HCl,..
Then DeltaH = Q/0.1

You wouldn't divide by 0.2 moles (or 0.2 + 0.1) ..Delta H is the heat released when 1 mole of Na2CO3 reacts with 2 moles of HCl...i.e. "per mole of equation"

It is a common point of confusion
Ah ok thanks again!
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