Calculate the standard enthalpy change for the net reaction from the following chemical equations and standard enthalpy changes.
A) 2C+O2-->2CO. =-221.0KJ
B) C+O2-->2CO2. = -393.5KJ
C) 2H2+O2-->2H2O. =-483.6KJ
NET: CO+H2+O2-->CO2+H2O =?
7 years ago
Answered By Melis G
Start by identifying the reactants (or products) in the net equation and choosing one of the equations above,
For example, CO
We need to pick equation A, however, the CO appears as a product(right side of arrow) and we need it to be on the reactant side as depicted in the net equation. Hence we need to flip equation A:
2CO --> 2C +O2 HA= + 221.0 kJ
Note that when you flip the equation, the sign will also flip (become positive in this case).
Proceed to identify the rest of the reactants and products and use the remaining equations to come up with the net equation. You can multiply or divide by factors as needed.
For example, we need to use equation C to get H2O as the product, but we need to divide by 2 so it matches the net equation. (Enthalpy will also be divided by 2).
When we apply a factor, all of the coefficients will change accordingly for that equation:
H2 + 1/2O2 --> H2O HC = -241.8 kJ
See if you can come up with the last equation yourself to get the final NET equation. At the end, you simply need to add all the enthalpies together (that you may or may not have applied a factor to).
7 years ago
Answered By Melis G
Start by identifying the reactants (or products) in the net equation and choosing one of the equations above,
For example, CO
We need to pick equation A, however, the CO appears as a product(right side of arrow) and we need it to be on the reactant side as depicted in the net equation. Hence we need to flip equation A:
2CO --> 2C +O2 HA= + 221.0 kJ
Note that when you flip the equation, the sign will also flip (become positive in this case).
Proceed to identify the rest of the reactants and products and use the remaining equations to come up with the net equation. You can multiply or divide by factors as needed.
For example, we need to use equation C to get H2O as the product, but we need to divide by 2 so it matches the net equation. (Enthalpy will also be divided by 2).
When we apply a factor, all of the coefficients will change accordingly for that equation:
H2 + 1/2O2 --> H2O HC = -241.8 kJ
See if you can come up with the last equation yourself to get the final NET equation. At the end, you simply need to add all the enthalpies together (that you may or may not have applied a factor to).