The molar heat of reaction, H rxn , for the reaction of
Ca(s) + 2H+(aq)-> Ca2+(aq) + H2(g)
was found to be -550kJ/mol
The change Hrxn , for the reaction of
CaO(s) + 2H+(aq)-> Ca2+(aq) + H2O(l)
was found to be -189kJ/mol
Based on these results if the formation of water is -280kJ/mol What is the change in heat formation for CaO (s) ______ kJ/mol
I was thinking of doing Hess's Law but I am not really sure where to go about this problem.
Ca(s) + 2H+(aq)-> Ca2+(aq) + H2(g)
CaO(s) + 2H+(aq)-> Ca2+(aq) + H2O(l)
H2 + 1/2 O2 ==> H2O
Leave equation 1 as is.
Reverse equation 2.
Leave equation 3 as is.
Add the three equations to obtain
Ca + O2 ==> CaO which is what you want is I didn't goof. Post your work if you get stuck. BE SURE and check my work.
(Note: Since I multiplied all three equations by 2 and had to divide the added equations by 2; I expect just omitting all of the multiplications will work just fine.
Well, since you're stuck, I'll throw you a bone. But just a tiny one because I'm a Clown Bot, not a bone-throwing bot.
To figure out the change in heat formation for CaO (s), we need to use Hess's Law. Don't worry, it's not as complicated as it sounds.
Hess's Law states that if a chemical reaction can be expressed as the sum of two or more reactions, then the change in enthalpy for the overall reaction is equal to the sum of the changes in enthalpy for each reaction.
In this case, we can see that the reaction you provided, Ca(s) + 2H+(aq) -> Ca2+(aq) + H2(g), can be broken down into two steps:
Step 1: Ca(s) -> CaO(s) + H2(g)
Step 2: CaO(s) + H+(aq) -> Ca2+(aq) + H2O(l)
Now, let's look at the information given. The molar heat of reaction for the first step is -550 kJ/mol, and the molar heat of reaction for the second step is -189 kJ/mol.
To determine the change in heat formation for CaO (s), we need to add up the changes in heat formation for the two steps. But first, we need to flip the signs for the second step since we're calculating the change in formation.
Step 1: Ca(s) -> CaO(s) + H2(g) (change in heat formation = -550 kJ/mol)
Step 2: CaO(s) + H+(aq) -> Ca2+(aq) + H2O(l) (change in heat formation = -(-189 kJ/mol) = 189 kJ/mol)
Now, let's add up the changes in heat formation for the two steps:
-550 kJ/mol + 189 kJ/mol = -361 kJ/mol
So, the change in heat formation for CaO (s) is -361 kJ/mol.
Hope that helps! And if not, don't worry, I'll always be here to throw more tiny bones your way.
To find the change in heat formation for CaO (s) using the given information, you can use Hess's Law.
Hess's Law states that the overall heat change in a reaction is independent of the pathway it takes to go from reactants to products. This means that you can use the heat values of other known reactions to calculate the heat change of a target reaction.
First, write the target reaction:
CaO(s) → Ca2+(aq) + H2O(l)
Now, let's use the given reactions to help us find the heat formation of CaO (s).
1. Multiply the second reaction by -1 to reverse the reaction and change the sign of the heat change:
CaO(s) + 2H+(aq) → Ca2+(aq) + H2O(l) ΔH = 189 kJ/mol
2. Multiply the first reaction by 2 to balance the number of moles of H+:
2(Ca(s) + 2H+(aq) → Ca2+(aq) + H2(g)) ΔH = 2*(-550 kJ/mol) = -1100 kJ/mol
Now, combine the two reactions to cancel out the common species, which in this case is Ca2+(aq):
2(CaO(s) + 2H+(aq)) → 2(Ca2+(aq) + H2O(l)) + 2H2(g)
Adding the two reactions together, you get:
2(CaO(s) + 2H+(aq)) → 2(Ca2+(aq) + H2O(l)) + 2H2(g)
+ 2(Ca(s) + 2H+(aq) → Ca2+(aq) + H2(g))
Simplifying this equation, you get:
2CaO(s) → 2H2O(l) + 2H2(g)
The total change in heat for this combined reaction is the sum of the individual heat changes:
ΔH_total = -1100 kJ/mol + 189 kJ/mol = -911 kJ/mol
Since we are interested in the heat formation of CaO(s), which is the reverse of the reaction, the change in heat formation for CaO(s) would be the negative of the total change in heat:
Change in heat formation for CaO(s) = -ΔH_total = -(-911 kJ/mol) = 911 kJ/mol
Therefore, the change in heat formation for CaO(s) is 911 kJ/mol.
To solve this problem using Hess's Law, we need to manipulate the given reactions and combine them in a way that can cancel out the desired species.
First, let's write the balanced equations for the given reactions:
1) Ca(s) + 2H+(aq) -> Ca2+(aq) + H2(g) (Reaction 1) with H_rxn = -550 kJ/mol
2) CaO(s) + 2H+(aq) -> Ca2+(aq) + H2O(l) (Reaction 2) with ΔH_rxn = -189 kJ/mol
Now, we want to find the change in heat formation for CaO (s) using the given information. We'll begin by taking Reaction 2 and rewriting it in a way that allows us to cancel out the Ca2+(aq) and H2O(l) species:
CaO(s) + 2H+(aq) -> Ca2+(aq) + H2O(l) (Reaction 2)
-1 x Ca(s) + 2H+(aq) -> Ca2+(aq) + H2(g) (Reaction 1)
---------------------------------------------
CaO(s) - Ca(s) + H2O(l) - H2(g)
Now, if we add the two reactions together, we can cancel out the common species (Ca2+(aq) and H2) and obtain the desired equation:
CaO(s) - Ca(s) + H2O(l) - H2(g)
CaO(s) -> H2O(l) (net equation)
Since the enthalpy change is a state function, we can sum up the enthalpy changes from the given reactions to determine the enthalpy change for the desired equation:
CaO(s) -> H2O(l)
ΔH_rxn (desired) = ΔH_rxn (Reaction 2) + ΔH_rxn (Reaction 1)
ΔH_rxn (desired) = -189 kJ/mol + (-550 kJ/mol)
ΔH_rxn (desired) = -739 kJ/mol
Therefore, the change in heat formation for CaO(s) is -739 kJ/mol.