I really need help with this I am really confused with the law, I have looked at multiple examples and I am still so confused, I know you can't give me the answer, but is there anyway you could please take me through this question step by step?
Use Hess's law and changes in enthalpy for the following to generic reactions to calculate (delta)H for the reaction 2A +2B2C3--> 2B +A2C3
2A + 2/3C2-->A2C3 (delta)H =-1874kJ
2B + 2/3C2--> B2C3 (delta)H=-285kJ
Thank you so much this would be so helpful to me.
Your two equations are not balanced and we can't do anything with them until they are.
For example, #1.
You have 2A on the left and 2A on the right. You have 2/3 C2 which is 4/3 C on the left but 3 on the right. Are you sure that isn't 2A + 3/2 C2 = A2C3?
And #2 would be 2B + 3/2 C2 = B2C3?
Of course, I can definitely walk you through this step by step. To calculate the change in enthalpy (ΔH) for the given reaction using Hess's law, we need to manipulate the given reactions to match the desired reaction.
Step 1: Reverse the second reaction
By reversing the second reaction, we obtain:
B2C3 --> 2B + 2/3C2, and ΔH becomes +285 kJ (since we reversed the reaction, we change the sign of ΔH).
Step 2: Double the first reaction
By doubling the first reaction, we get:
4A + 4/3C2 --> 2A2C3, and ΔH remains -1874 kJ (as the coefficients are doubled, ΔH remains the same).
Step 3: Cancel out common compounds
Now, we can add the manipulated reactions together to obtain the desired reaction:
4A + 4/3C2 + B2C3 --> 2A2C3 + 2B + 2/3C2
To do this, we need to cancel out the common compounds on both sides of the equation. In this case, we can cancel out:
4/3C2
After canceling out, we are left with:
4A + B2C3 --> 2A2C3 + 2B
Step 4: Calculate the overall ΔH
To find the overall ΔH for the desired reaction, we add the ΔH values of the manipulated reactions:
ΔH = (-1874 kJ) + (+285 kJ)
ΔH = -1589 kJ
Therefore, the change in enthalpy (ΔH) for the reaction:
2A + 2B2C3 --> 2B + A2C3 is equal to -1589 kJ.
I hope this step-by-step explanation helps you understand how to use Hess's law to calculate the change in enthalpy for a given reaction. Let me know if you have any further questions!
Certainly! I can guide you through the process step by step.
To use Hess's law, we need to manipulate the given reactions in order to obtain the target reaction. Here's how we can do it:
Step 1: Write the given reactions and their enthalpy changes:
Reaction 1: 2A + 2/3C2 → A2C3 ΔH = -1874 kJ
Reaction 2: 2B + 2/3C2 → B2C3 ΔH = -285 kJ
Step 2: Compare the given reactions to the target reaction and identify the necessary steps to reach it.
In the target reaction, we need to obtain 2A and A2C3 on the reactant side, and 2B and B2C3 on the product side.
Step 3: Reverse one of the reactions if necessary.
In reaction 1, A2C3 is on the product side, but we need it on the reactant side. So, we reverse it.
Revised Reaction 1: A2C3 → 2A + 2/3C2 ΔH = +1874 kJ
Step 4: Multiply or divide the reactions to match the stoichiometry of the target reaction.
In reaction 2, we need 2B on the product side, but currently, we only have one. So, we multiply reaction 2 by 2.
Revised Reaction 2: 4B + 4/3C2 → 2B2C3 ΔH = -570 kJ
Step 5: Combine the manipulated reactions to obtain the target reaction.
When we add the revised reactions together, the reactants on one side will cancel out the products on the other side, leaving us with the target reaction.
Target Reaction: 2A + 2B2C3 → 2B + A2C3
Now, we need to sum up the enthalpy changes to calculate ΔH for the target reaction.
Step 6: Add up the enthalpy changes.
-1874 kJ + (-570 kJ) = -2444 kJ
So, the ΔH for the given reaction 2A + 2B2C3 → 2B + A2C3 is -2444 kJ.
I hope this step-by-step explanation helps you understand how to use Hess's law to calculate ΔH for a reaction.