Calculate the enthalpy change for the reaction 2C (s) + H2 (g) yield C2H2 (g) given the following reactions and their respective enthalpy changes:
C2H2(g) + 5/2 O2(g) yield 2CO2(g) + H2O (l) = -1299.6kJ
C(s) + O2(g) yield CO2 (g) -393.5
H2(g) + 1/2 O2(g) yield H2O(l) -285.9
Would the answer be 226.73kJ
I showed you how to do this a few days ago but I don't know if you saw my suggestions for working the problem. Yes, the answer is 226.7 kJ. Good work.
To calculate the enthalpy change for the given reaction, we can use Hess's law. Hess's law states that the total enthalpy change for a reaction is the sum of the enthalpy changes of the individual reactions that make up the overall reaction.
First, let's write the balanced chemical equation for the given reaction:
2C(s) + H2(g) → C2H2(g)
Now, let's manipulate the given equations in order to cancel out the necessary species and obtain the desired overall reaction:
1. Multiply the second equation by 2 to match the coefficient of C2H2:
2C(s) + 2O2(g) → 2CO2(g)
2. Multiply the third equation by 2 and reverse it to match the coefficient of H2:
H2(g) + O2(g) → 2H2O(l)
Now, we add up the manipulated equations:
2C(s) + 2O2(g) + H2(g) + O2(g) → 2CO2(g) + 2H2O(l)
Since the equation above contains all the necessary species for the desired reaction, we can sum up the enthalpy changes of the manipulated equations to find the enthalpy change for the overall reaction:
ΔH = ΔH1 + ΔH2 + ΔH3
ΔH = (-1299.6 kJ) + (-393.5 kJ) + (-2 * (-285.9 kJ))
ΔH = -1299.6 kJ - 393.5 kJ + 571.8 kJ
ΔH = -1121.3 kJ + 571.8 kJ
ΔH = -549.5 kJ
Therefore, the enthalpy change for the reaction 2C(s) + H2(g) → C2H2(g) is -549.5 kJ, not 226.73 kJ.
To calculate the enthalpy change for the reaction 2C (s) + H2 (g) → C2H2 (g), we can use Hess's Law.
First, we need to flip and adjust the given equations to match the reaction we are interested in:
C2H2(g) + 5/2 O2(g) → 2CO2(g) + H2O (l) ΔH = -1299.6 kJ
C(s) + O2(g) → CO2 (g) ΔH = -393.5 kJ
H2(g) + 1/2 O2(g) → H2O(l) ΔH = -285.9 kJ
Next, we need to manipulate these equations to obtain the target reaction: 2C (s) + H2 (g) → C2H2 (g).
Multiply the second equation by 2 to get:
2C(s) + 2O2(g) → 2CO2(g) ΔH = -787 kJ
Multiply the third equation by 2 to get:
2H2(g) + O2(g) → 2H2O(l) ΔH = -571.8 kJ
Now, we can add up these modified equations to obtain the target reaction:
2C(s) + 2H2(g) + 2O2(g) → C2H2(g) + 2CO2(g) + 2H2O(l)
The enthalpy change for this reaction is the sum of the ΔH values from the individual equations:
ΔH = (-787 kJ) + (-571.8 kJ) = -1358.8 kJ
Therefore, the enthalpy change for the reaction 2C(s) + H2(g) → C2H2(g) is -1358.8 kJ.
So, the calculated enthalpy change is -1358.8 kJ, not 226.73 kJ.