help me in chem

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Hedwigeeeee
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#1
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why when we calculate the standard enthalpy of neutralisation, we should use -Q/n, which n is the amount of acid , why not base?
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Mad_Dog_Graveson
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#2
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Mix Francium with Hydrochloric Acid
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Hedwigeeeee
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#3
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(Original post by Mad_Dog_Graveson)
Mix Francium with Hydrochloric Acid
Is that means as acid is neutralised by base, so we should divide the enthalpy change by the amount of acid to obtain standard enthalpy change of neutralization? why n is not amount of acid plus amount of base, but acid only? sorry i did not get your point...
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Pigster
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(Original post by Wuxinyu)
why when we calculate the standard enthalpy of neutralisation, we should use -Q/n, which n is the amount of acid , why not base?
n is the amount of water produced, not the amount the acid OR the amount of base.
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Kian Stevens
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(Original post by Wuxinyu)
why when we calculate the standard enthalpy of neutralisation, we should use -Q/n, which n is the amount of acid , why not base?
Consider that in any acid-base neutralisation, H+ + OH- \rightarrow H2O

Thus, depending on what the limiting reagent is, the moles of acid or base will be equivalent to the moles of water produced; thus, n is the moles of water produced as Pigster said, and this is because the enthalpy of neutralisation is defined in terms of the water produced
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Hedwigeeeee
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#6
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(Original post by Kian Stevens)
Consider that in any acid-base neutralisation, H+ + OH- \rightarrow H2O

Thus, depending on what the limiting reagent is, the moles of acid or base will be equivalent to the moles of water produced; thus, n is the moles of water produced as Pigster said, and this is because the enthalpy of neutralisation is defined in terms of the water produced
but when the amount of acid and base are not equal, they are not equivalent to the amount of water produced, the question in the attachment does not have the same volume of HCL and NaOH. they used the amount of base, which made me confused.
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Kian Stevens
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(Original post by Wuxinyu)
but when the amount of acid and base are not equal, they are not equivalent to the amount of water produced, the question in the attachment does not have the same volume of HCL and NaOH. they used the amount of base, which made me confused.
Question 1b even states that you have to calculate the enthalpy change per mole of water produced, so this gives you a hint anyway as this is what the enthalpy change of neutralisation is essentially measured in; don't forget that the enthalpy change of neutralisation is defined as the enthalpy change of an acid-base reaction which produces one mole of water

So, like I said, this is dictated by what the limiting reagent is; you don't always use the moles of acid, as the limiting reagent could be base, and based on the neutralisation ionic equation I stated earlier the moles of limiting reagent used will equal the moles of water produced... This is what you use as n
Last edited by Kian Stevens; 2 years ago
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Hedwigeeeee
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#8
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(Original post by Kian Stevens)
Question 1b even states that you have to calculate the enthalpy change per mole of water produced, so this gives you a hint anyway as this is what the enthalpy change of neutralisation is essentially measured in; don't forget that the enthalpy change of neutralisation is defined as the enthalpy change of an acid-base reaction which produces one mole of water

So, like I said, this is dictated by what the limiting reagent is; you don't always use the moles of acid, as the limiting reagent could be base, and based on the neutralisation ionic equation I stated earlier the moles of limiting reagent used will equal the moles of water produced... This is what you use as n
sorry, could you explain what is limiting reagent?
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