Calculate ∆H for the reaction
No(g) + 1/2 O2(g) ----> NO2(g)
if all the products and reactants are under standard conditions. NO and NO2 are formed from their elements according to following equations
1/2 N2 + 1/2 O2(g) ---> NO(g) +90.4 kJ/mol
1/2 N2(g) +O2(g) ---> NO2(g) +33.9 kJ/mol
answer is units of kJ/mol rxn
I rather confused. which equations do i have to flip? what do i have to multiply them by?
No(g) + 1/2 O2(g) ----> NO2(g)
if all the products and reactants are under standard conditions. NO and NO2 are formed from their elements according to following equations
1/2 N2 + 1/2 O2(g) ---> NO(g) +90.4 kJ/mol
1/2 N2(g) +O2(g) ---> NO2(g) +33.9 kJ/mol
answer is units of kJ/mol rxn
I rather confused. which equations do i have to flip? what do i have to multiply them by?
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For this problem, you will need to flip the equation
1/2 N2 + 1/2 O2 ---> NO .
You need to flip this one because the NO is on the left side of the equation.
There is no need to multiply by anything in this case because there is sufficient amount of ever compound that is needed
Hope this helped :)
1/2 N2 + 1/2 O2 ---> NO .
You need to flip this one because the NO is on the left side of the equation.
There is no need to multiply by anything in this case because there is sufficient amount of ever compound that is needed
Hope this helped :)