exothermic reaction

[inaccensa]

For a given exothermic reaction, how are ΔH, Ea (forward) and Ea (reverse) related?
A. ΔH + Ea (forward) = Ea (reverse)
B. ΔH - Ea (forward) = Ea (reverse)
C. ΔH = Ea (forward) + Ea (reverse)
D. ΔH = Ea (forward) - Ea (reverse)

change in enthalpy = E reverse - E forward
ΔH + Ea (forward) = Ea (reverse) is the answer


but I was just wondering whether we put a negative sign before the enthalpy since its given as an exothermic reaction

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[crazybob]

i don't think the sign is necessary for delta H. the value of it is what will determine the sign.

 

[pnoybballin]

Wait, could you explain why the answer is.. ΔH + Ea (forward) = Ea (reverse) is the answer? I thought for exothermic reactions, heat is released when bonds are formed and heat is used to break bonds thus Ea (forward) = Ea + ΔH (reverse), so wouldn't the answer be D. ΔH = Ea (forward) - Ea (reverse)?

 

[crazybob]

let's consider three energy states

• R = reactants
• T = transition state
• P = products

Then,

• Ea (forward) = T - R
• Ea (reverse) = T - P
• delta H = P - R (because change = final - initial)

If we find expressions for P and R, we can find delta H

• R = T - Ea (forward)
• P = T - Ea (reverse)

So, delta H = P - R = T - Ea (forward) - (T - Ea (reverse) )
= T - Ea (forward) - T + Ea (reverse)
= -Ea (forward) + Ea (reverse)
delta H = Ea (reverse) - Ea (forward)

The correct answer is A.

Btw, if you look at choices B and C, you realize they're both the same expressions.

 

[Entadus]

Wait, could you explain why the answer is.. ΔH + Ea (forward) = Ea (reverse) is the answer? I thought for exothermic reactions, heat is released when bonds are formed and heat is used to break bonds thus Ea (forward) = Ea + ΔH (reverse), so wouldn't the answer be D. ΔH = Ea (forward) - Ea (reverse)?

Agreed. Kaplan got this incorrect, I think

let's consider three energy states

• R = reactants
• T = transition state
• P = products

Then,

• Ea (forward) = T - R
• Ea (reverse) = T - P
• delta H = P - R (because change = final - initial)

If we find expressions for P and R, we can find delta H

• R = T - Ea (forward)
• P = T - Ea (reverse)

So, delta H = P - R = T - Ea (forward) - (T - Ea (reverse) ) No this is wrong, it should be flipped: P - R = T - Ea (reverse) - (T - Ea (forward) )
= T - Ea (forward) - T + Ea (reverse) = T - Ea (reverse) - T + Ea (forward)
= -Ea (forward) + Ea (reverse) = Ea (forward) - Ea (reverse)
delta H = Ea (reverse) - Ea (forward) delta H = Ea (forward) - Ea (reverse)

The correct answer is A. The correct answer is therefore D

Btw, if you look at choices B and C, you realize they're both the same expressions.

There is a mistake made (bold corrections are mine)

Ea (reverse) is bigger than Ea (forward), so they needed to be subtracted in the correct order to get a negative delta H (Exothermic)

 

[Captain Sisko]

Agreed. Kaplan got this incorrect, I think



There is a mistake made (bold corrections are mine)

Ea (reverse) is bigger than Ea (forward), so they needed to be subtracted in the correct order to get a negative delta H (Exothermic)

cool. i got d.