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Chemistry Enthalpy Change OCR A

Could anyone please explain how to do question 5bi and 5bii please, thanks. It is also on page 12.

https://pmt.physicsandmathstutor.com/download/Chemistry/A-level/Topic-Qs/OCR-A/3-Periodic-Table-and-Energy/Set-F/Enthalpy%20Changes%201%20QP.pdf

Reply 1

TypicalNerd

Original post by koketenshi

Could anyone please explain how to do question 5bi and 5bii please, thanks. It is also on page 12.
https://pmt.physicsandmathstutor.com/download/Chemistry/A-level/Topic-Qs/OCR-A/3-Periodic-Table-and-Energy/Set-F/Enthalpy%20Changes%201%20QP.pdf

5(b)(i) - look carefully at the reaction profile. You can ignore the arrow going from the reactants level to the top of the hump as this is the activation energy.

The enthalpy change is the other arrow, labelled +164 kJ/mol. If you look at the products level of the reaction profile, 2 mol of N2O are made and so this is the enthalpy change when 2 mol of N2O are made.

Recall that at RTP 1 mole of gas occupies 24 dm^3, so 240 dm^3 of gas is 10 moles. This is five times the amount the reaction profile shows forming and so your answer is 5 x +164 kJ = +820 kJ.

5(b)(ii) - As above, the enthalpy change when 2 mol of N2O (g) are made is given. The standard enthalpy change of formation is defined to be that when 1 mole of product is formed from its constituent elements with everything in its standard state. As such, you just halve the given value and should get +82 kJ/mol.

(edited 1 month ago)

Reply 2

dsdf_gfdgfdgfd

Original post by koketenshi

Could anyone please explain how to do question 5bi and 5bii please, thanks. It is also on page 12.
https://pmt.physicsandmathstutor.com/download/Chemistry/A-level/Topic-Qs/OCR-A/3-Periodic-Table-and-Energy/Set-F/Enthalpy%20Changes%201%20QP.pdf

where did u get these questions from

Reply 3

Pigster

Original post by dsdf_gfdgfdgfd

where did u get these questions from

The web address literally says "physicsandmathstutor.com"

Unless you meant where did the Qs originate from, then the answer is the previous OCR A spec past papers.

(edited 1 month ago)

Reply 4

koketenshi

Original post by dsdf_gfdgfdgfd

where did u get these questions from

pmt - physics and maths tutor

(edited 1 month ago)

Reply 5

koketenshi

Original post by TypicalNerd

5(b)(i) - look carefully at the reaction profile. You can ignore the arrow going from the reactants level to the top of the hump as this is the activation energy.
The enthalpy change is the other arrow, labelled +164 kJ/mol. If you look at the products level of the reaction profile, 2 mol of N2O are made and so this is the enthalpy change when 2 mol of N2O are made.
Recall that at RTP 1 mole of gas occupies 24 dm^3, so 240 dm^3 of gas is 10 moles. This is five times the amount the reaction profile shows forming and so your answer is 5 x +164 kJ = +820 kJ.
5(b)(ii) - As above, the enthalpy change when 2 mol of N2O (g) are made is given. The standard enthalpy change of formation is defined to be that when 1 mole of product is formed from its constituent elements with everything in its standard state. As such, you just halve the given value and should get +82 kJ/mol.

thank you, that was rly helpful and if you don't mind could you also explain question 3b) and 3c) on page 7 please.

(edited 1 month ago)

Reply 6

TypicalNerd

Original post by koketenshi

thank you, that was rly helpful and if you don't mind could you also explain question 3b) and 3c) on page 7 please.

3(b)(i) - The activation energy of the reverse reaction is equal to the activation energy of the forward reaction plus the enthalpy change of the reverse reaction. This is perhaps easiest to see when you look at the completed reaction profile (the activation energy is always the height of the top of the hump above the level of interest - for the reverse reaction, the level of interest is always the line representing the products of the reaction).

The reverse reaction must have an enthalpy change of +9 kJ/mol because the forward reaction has an enthalpy change of -9 kJ/mol (the enthalpy change of the forward reaction is always the negative of the enthalpy change of the reverse reaction) and we are told the activation energy of the forward reaction is +173 kJ/mol. Adding these up, we should get +182 kJ/mol.

3(c) - The strategy here is similar to the one used in 5(b)(ii), but there is a small twist in that the question now asks about the reverse reaction instead of the forward reaction (e.g 2HI -> H2 + I2) and so the enthalpy change is +9 kJ/mol and not -9 kJ/mol. Furthermore, this equation features two moles of HI rather than one and so some care needs to be taken when picking a suitable number to scale the enthalpy change up by.

Since it tells you the gas is measured at RTP, we know each mole of gas occupies 24 dm^3 and so the 336 dm^3 of HI must contain 14 moles. This is seven times the number of moles of HI used in the equation, you multiply the enthalpy change of the reverse reaction by 7. That is to say the enthalpy change is +63 kJ.