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Chemistry Question 128 – AP-EAMCET 2026

Consider the following reaction A2B4(g)2AB2(g) At 300 K, ΔrH for this reaction is x kJ mol1. What is its ΔrU (in kJ mol1) at the same temperature? (R=8.3 J mol1 K1)

The relationship between enthalpy (ΔH) and internal energy (ΔU) depends on the change in the number of moles of gaseous substances in a reaction.

Step 1: Determine the change in moles of gas (Δng)✦ Active

The given reaction is A2B4(g)2AB2(g). From the balanced chemical equation, we can find the number of moles of gaseous products and reactants.

Moles of gaseous products=2 (from 2AB2(g)) Moles of gaseous reactants=1 (from A2B4(g)) Δng=(moles of gaseous products)(moles of gaseous reactants) Δng=21=1 mol
💡 Teacher's Secret Hint

Remember to only consider gaseous species when calculating Δng.

Step 2: State the relationship between ΔH and ΔU and identify given values○ Expand

The relationship between change in enthalpy (ΔH) and change in internal energy (ΔU) is given by:

ΔH=ΔU+ΔngRT

We are given: ΔrH=x kJ mol1 T=300 K R=8.3 J mol1 K1 And we found Δng=1 mol.

💡 Teacher's Secret Hint

Ensure all units are consistent. Since ΔH is in kJ, it's usually best to convert J to kJ for the ΔngRT term.

Step 3: Calculate the ΔngRT term and convert units○ Expand

Now, calculate the product ΔngRT:

ΔngRT=(1 mol)×(8.3 J mol1 K1)×(300 K) ΔngRT=2490 J

Since ΔH is given in kJ, convert ΔngRT from Joules to kilojoules (1 kJ=1000 J):

ΔngRT=24901000 kJ=2.49 kJ
💡 Teacher's Secret Hint

A common mistake is forgetting to convert units, especially between Joules and kilojoules. Always double-check unit consistency before final calculations.

Step 4: Solve for ΔrU○ Expand

Rearrange the formula ΔH=ΔU+ΔngRT to solve for ΔU:

ΔU=ΔHΔngRT

Substitute the given value of ΔrH=x kJ mol1 and the calculated value of ΔngRT=2.49 kJ mol1:

ΔrU=(x kJ mol1)(2.49 kJ mol1) ΔrU=(x+2.49) kJ mol1

Comparing this result with the given options, it matches option 3.

💡 Teacher's Secret Hint

Pay close attention to the signs when substituting values. A negative ΔH and a positive ΔngRT can sometimes lead to different interpretations of the final sign.

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