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H2 + 1/2 O2 —> H2O -286kJ
N2O5 + H2O —> 2HNO3 -77kJ
1/2N2 + 3/2 O2 + 1/2 H2 —> HNO3 -174kJ
Calculate the *Delta* H for
2N2 + 5O2 —-> 2N2O5
I attempted this problem and got +20kJ. Was I right?
multiply 3rd eqn by 4
... 2 N2 + 6 O2 + 2 H2 —> 4 HNO3 -696kJ
reverse 2nd eqn and multiply by 2
... 4 HNO3 —> 2 N2O5 + 2 H2O +154kJ
reverse 1st eqn and multiply by 2
... 2 H2O —> 2 H2 + O2 +572Kj
add the the three eqn, cancel identical items on both sides
-696 + 154 + 572 = ? ... check your math
To calculate the ∆H for a reaction using Hess's Law, you need to determine the correct combination and manipulation of the given reactions to obtain the desired overall reaction. Let's break down the process step by step:
Step 1: Write the given reactions and their corresponding enthalpy changes:
1. H2 + 1/2 O2 —> H2O ∆H = -286 kJ (Reaction 1)
2. N2O5 + H2O —> 2HNO3 ∆H = -77 kJ (Reaction 2)
3. 1/2 N2 + 3/2 O2 + 1/2 H2 —> HNO3 ∆H = -174 kJ (Reaction 3)
Step 2: Adjust the coefficients and manipulations of the given reactions to match the desired overall reaction:
Given overall reaction: 2N2 + 5O2 —> 2N2O5
To accomplish this, let's manipulate the given reactions to obtain the desired reaction:
a) Multiply Reaction 1 by 2 to obtain 2H2 + O2 —> 2H2O
b) Multiply Reaction 2 by 2 to obtain 2N2O5 + 2H2O —> 4HNO3
c) Reverse Reaction 3 to obtain HNO3 —> 1/2 N2 + 3/2 O2 + 1/2 H2
So, the adjusted reactions are:
1. 2H2 + O2 —> 2H2O, with an adjusted ∆H = -572 kJ
2. 2N2O5 + 2H2O —> 4HNO3, with an adjusted ∆H = -154 kJ
3. HNO3 —> 1/2 N2 + 3/2 O2 + 1/2 H2, with an adjusted ∆H = 174 kJ
Step 3: Add the adjusted reactions to obtain the desired reaction:
2N2 + 5O2 —> 2N2O5
Adding Reaction 2 and Reaction 3 cancels out HNO3 and gives:
2N2O5 + 2H2O + HNO3 —> 4HNO3 + 1/2 N2 + 3/2 O2 + 1/2 H2
Subtracting Reaction 1 from the reaction above gives:
2N2O5 + 2H2O + HNO3 - 2H2 - O2 —> 2N2O5 + 1/2 N2 + 3/2 O2 + 1/2 H2
Simplifying the equation gives:
2N2O5 - 2H2 - O2 —> 1/2 N2 + 3/2 O2 + 1/2 H2
The adjusted overall reaction has an adjusted ∆H = -572 kJ + 174 kJ = -398 kJ.
Therefore, the ∆H for the reaction 2N2 + 5O2 —> 2N2O5 is -398 kJ.
Based on your calculation, where you obtained a ∆H of +20 kJ, it seems there might have been an error. Make sure to check your manipulation of the given reactions and the algebraic operations performed.
It's crucial to double-check the steps and calculations to ensure accuracy when using Hess's Law to determine the enthalpy change for a reaction.