From the statement of Hess' law, the enthalpy of the reaction A---> C is +90 kJ
<h3>What is Hess' law?</h3>
Hess' law of constant heat summation states that for a multistep reaction, the standard enthalpy of reaction is always constant and is independent of the pathway or intermediate routes taken.
From Hess' law, the enthalpy change for the reaction A ----> C is calculated as follows:
A---> C = A ---> B + B ---> C
ΔH of A---> C = 30 kJ + 60 kJ
ΔH = 90 kJ
Therefore, the enthalpy of the reaction A---> C is +90 kJ
The above reaction A---> C can be shown in the enthalpy diagram below:
A -------------------> C (ΔH = +90 kJ)
\ /
\ / (ΔH = +60 kJ)
(ΔH = +30 J) \ /
> B
Learn more about enthalpy and Hess law at: brainly.com/question/9328637
Answer:
Explanation:
Catalyst is I2 . Because I2 is reacted with starting material in step 1 and generated in second step
Rate limiting step is step 1. Because in rate equation CH3CHO and I2 is mentioned. Hence the overall rate of reaction is depending CH3CHO and I2. Rate limiting step is step 1
