Ok, I really need to know how to work problems like these so could someone please show me how to set it up?! Given this data:
1. C(graphite) + O2(g) produces CO2(g), delta H = -393.5 kJ
2. H2(g) + 1/2 O2(g) produces H2O(l), delta H = -285.8 kJ
3. Ch3OH(l) + 3/2 O2(g) produces Co2(g) + 2H2O(l), delta H = -726.4 kJ
Given these calculate the enthalpy change for:
C(graphite) + 2H2(g) + 1/2 O2(g) produces CH3OH(l)Please help!
Use equation 1 as is. (delta H as is)
Add 2x equation 2 (and 2x delta H).
Reverse equation 3 and add to the other two (change the sign on delta H).
Then added the equations as noted and confirm you obtain the desired equation; finally, add the delta Hs (with changes as noted).
To calculate the enthalpy change for the reaction C(graphite) + 2H2(g) + 1/2 O2(g) produces CH3OH(l), we can use Hess's Law, which states that the enthalpy change of a reaction is independent of the pathway between the initial and final states but depends only on the initial and final states.
Here's how you can set up the problem:
1. Write down the given reactions and their corresponding enthalpy changes:
a. C(graphite) + O2(g) produces CO2(g), delta H = -393.5 kJ
b. H2(g) + 1/2 O2(g) produces H2O(l), delta H = -285.8 kJ
c. Ch3OH(l) + 3/2 O2(g) produces Co2(g) + 2H2O(l), delta H = -726.4 kJ
2. Rearrange the given reactions to match the desired reaction:
a. CO2(g) produces C(graphite) + O2(g), delta H = +393.5 kJ (reversed sign)
b. H2O(l) produces H2(g) + 1/2 O2(g), delta H = +285.8 kJ (reversed sign)
c. Co2(g) + 2H2O(l) produces Ch3OH(l) + 3/2 O2(g), delta H = +726.4 kJ (reversed sign)
3. Multiply the coefficients of each reaction and adjust their enthalpy changes accordingly to ensure the same stoichiometry for the desired reaction:
a. 2 * (C(graphite) + O2(g) produces CO2(g)), delta H = 2 * (-393.5 kJ) = -787.0 kJ
b. 3 * (H2(g) + 1/2 O2(g) produces H2O(l)), delta H = 3 * (-285.8 kJ) = -857.4 kJ
c. 2 * (Ch3OH(l) + 3/2 O2(g) produces Co2(g) + 2H2O(l)), delta H = 2 * (-726.4 kJ) = -1452.8 kJ
4. Add up the modified reactions to obtain the desired reaction and its enthalpy change:
Final reaction: C(graphite) + 2H2(g) + 1/2 O2(g) produces CH3OH(l)
Final enthalpy change = (-787.0 kJ) + (-857.4 kJ) + (-1452.8 kJ) = -3097.2 kJ
Therefore, the enthalpy change for the reaction C(graphite) + 2H2(g) + 1/2 O2(g) produces CH3OH(l) is -3097.2 kJ.