Given the following data:
MnO(s) + H2(g) Mn(s) + H2O(l), H = 100 kJ
MnO2(s) + 2 H2(g) Mn(s) + 2 H2O(l), H = –52 kJ
use Hess's law to calculate the heat of the reaction
MnO2(s) + H2(g) MnO(s) + H2O(l)
Given the following data:
MnO(s) + H2(g) Mn(s) + H2O(l), H = 100 kJ
MnO2(s) + 2 H2(g) Mn(s) + 2 H2O(l), H = –52 kJ
use Hess's law to calculate the heat of the reaction
MnO2(s) + H2(g) MnO(s) + H2O(l)
Ah yes, Hess's law, the law that sounds like it belongs in a chemistry courtroom. Don't worry, I'll help you crack this case!
We need to use the given reactions to calculate the heat of the desired reaction. Let's call the desired reaction "X". We can break it down into smaller steps that we have heat values for.
Step 1: MnO2(s) + 2 H2(g) Mn(s) + 2 H2O(l) (given reaction 2)
Step 2: MnO2(s) + H2(g) MnO(s) + H2O(l) (desired reaction)
Step 3: Mn(s) + H2O(l) MnO(s) + H2(g) (reverse of given reaction 1)
Now, let's assign heat values based on the given reactions:
Heat of reaction 2: -52 kJ
Heat of reaction 1: 100 kJ
To calculate the heat of the desired reaction X, we need to take into account the reversibility of reaction 3 and its heat value:
Heat of reaction 3: -100 kJ (negative because it's the reverse reaction of given reaction 1)
Now, let's add up the heat values of the steps to get the heat of reaction X:
Step 1: -52 kJ
Step 2: unknown, let's call it "X"
Step 3: -100 kJ
-52 kJ + X - 100 kJ = unknown heat value
To find the unknown heat value, we can simply add the heat values on both sides:
X = -52 kJ + 100 kJ
Solving this (mathematical) mystery gives us:
X = 48 kJ
So, the heat of the "MnO2(s) + H2(g) MnO(s) + H2O(l)" reaction is 48 kJ. Case closed!
To calculate the heat of the reaction MnO2(s) + H2(g) -> MnO(s) + H2O(l), we can use Hess's law and the given data.
Hess's law states that the total enthalpy change of a reaction is independent of the pathway taken, as long as the initial and final conditions are the same.
First, let's analyze the given reactions and see how we can combine them to obtain the desired reaction:
1) MnO(s) + H2(g) -> Mn(s) + H2O(l): H = +100 kJ
2) MnO2(s) + 2 H2(g) -> Mn(s) + 2 H2O(l): H = -52 kJ
We can see that the reactants and products in reaction 2) are the same as the desired reaction, but the coefficients are different. To align the coefficients, we can multiply reaction 1) by 2, so that the number of moles of H2 matches:
2 MnO(s) + 2 H2(g) -> 2 Mn(s) + 2 H2O(l): ΔH = 2 * (+100 kJ) = +200 kJ
Now, we can reverse reaction 2) to match the reactants in the desired reaction:
Mn(s) + 2 H2O(l) -> MnO2(s) + 2 H2(g): ΔH = +52 kJ
Finally, we can add the two equations together to cancel out Mn and H2O:
2 MnO(s) + 2 H2(g) + Mn(s) + 2 H2O(l) -> MnO2(s) + 2 H2(g) + Mn(s) + 2 H2O(l)
Simplifying, we get:
MnO(s) + Mn(s) -> MnO2(s)
Now, since the overall reaction equation has canceled out the H2(g) and H2O(l), we can calculate the heat of the reaction using Hess's law:
ΔH = ΔH2 - ΔH1
ΔH = (-52 kJ) - (+200 kJ)
ΔH = -52 kJ - 200 kJ
ΔH = -252 kJ
Therefore, the heat of the reaction MnO2(s) + H2(g) -> MnO(s) + H2O(l) is -252 kJ.
To calculate the heat of the reaction MnO2(s) + H2(g) → MnO(s) + H2O(l) using Hess's law, we can use the given reactions and their respective enthalpy changes.
First, let's write the balanced equation for the target reaction:
MnO2(s) + H2(g) → MnO(s) + H2O(l)
Now, we need to manipulate the given reactions so that the desired reaction is obtained. We can reverse the first reaction and multiply the second reaction to achieve the desired reactants and products:
2(MnO(s) + H2O(l)) → 2Mn(s) + 2H2O(l) (By reversing the first reaction)
2(MnO2(s) + 2H2(g)) → 2MnO(s) + 2H2O(l) (Multiplying the second reaction by 2)
Next, we can add these manipulated reactions together:
2(MnO(s) + H2O(l)) + 2(MnO2(s) + 2H2(g)) → 2Mn(s) + 2H2O(l) + 2MnO(s) + 2H2O(l)
Simplifying the equation further gives:
2MnO(s) + 2 H2(g) + MnO2(s) + 4 H2(g) → 2Mn(s) + 5H2O(l) + MnO(s)
Now, we can calculate the resulting enthalpy change for this combined reaction by summing up the individual enthalpy changes:
2 × (Enthalpy change of MnO(s) + H2O(l)) + 2 × (Enthalpy change of MnO2(s) + 2H2(g)) = Enthalpy change of the target reaction.
Using the given enthalpy changes:
2 × (-100 kJ/mol) + 2 × (-52 kJ/mol) = Enthalpy change of the target reaction
Simplifying the expression:
-200 kJ/mol + (-104 kJ/mol) = Enthalpy change of the target reaction
Therefore, the heat of the reaction MnO2(s) + H2(g) → MnO(s) + H2O(l) is -304 kJ.