$CO_{(1)}+2 H_{2(\,\theta)} \rightarrow CH_3 OH_{(0)}$
Using the following data, what would be the enthalpy change for the reaction?
(i) $C _{(s)}+\frac{1}{2} O _{2(s)} \rightarrow CO _{(s)}$
$\Delta H ^0=-110.5 kJ / mol$
(ii) $C _{(\text {a })}+ O _{2(\text { g })} \rightarrow CO _{2(\text { g })}$
$\Delta H ^0=-393.5 kJ / mol$
(iii) $H _{2(g)}+\frac{1}{2} O _{2(s)} \rightarrow H _2 O _{( i )}$,
$\Delta H ^0=-285.5 kJ / mol$
(iv) $CH _3 OH _{(1)}+\frac{3}{2} O _{2(\beta)} \rightarrow 2 H _2 O _{(\beta)}+ CO _{2(\beta)}$
$\Delta H ^0=-726.6 kJ / mol$
Answer (2)
Reverse reaction (iv) to get 2 H 2 O ( l ) + C O 2 ( g ) → C H 3 O H ( l ) + 2 3 O 2 ( g ) with Δ H − 4 0 = 726.6 kJ/mol .
Add reaction (i), 2 times reaction (iii), and the reversed reaction (iv) to obtain the target reaction.
Calculate the enthalpy change: Δ H 0 = Δ H 1 0 + 2Δ H 3 0 + Δ H − 4 0 .
The enthalpy change for the reaction is 45.1 kJ/mol .
Explanation
Analyze the problem and available data We are given the following reactions and their enthalpy changes:
C ( s ) + 2 1 O 2 ( g ) → C O ( g ) , Δ H 1 0 = − 110.5 kJ/mol
C ( s ) + O 2 ( g ) → C O 2 ( g ) , Δ H 2 0 = − 393.5 kJ/mol
H 2 ( g ) + 2 1 O 2 ( g ) → H 2 O ( l ) , Δ H 3 0 = − 285.5 kJ/mol
C H 3 O H ( l ) + 2 3 O 2 ( g ) → 2 H 2 O ( l ) + C O 2 ( g ) , Δ H 4 0 = − 726.6 kJ/mol
Our target reaction is:
C O ( g ) + 2 H 2 ( g ) → C H 3 O H ( l )
We need to manipulate the given reactions to obtain the target reaction and then calculate the enthalpy change for the target reaction using Hess's Law.
Manipulate the equations First, we reverse reaction (4):
2 H 2 O ( l ) + C O 2 ( g ) → C H 3 O H ( l ) + 2 3 O 2 ( g ) , Δ H − 4 0 = 726.6 kJ/mol
Next, we add reaction (1), 2 times reaction (3), and the reversed reaction (4):
C ( s ) + 2 1 O 2 ( g ) + 2 ( H 2 ( g ) + 2 1 O 2 ( g ) ) + 2 H 2 O ( l ) + C O 2 ( g ) → C O ( g ) + 2 H 2 O ( l ) + C H 3 O H ( l ) + 2 3 O 2 ( g ) + C ( s ) + O 2 ( g )
Simplifying the equation, we obtain the target reaction:
C O ( g ) + 2 H 2 ( g ) → C H 3 O H ( l )
Calculate the enthalpy change Now, we calculate the enthalpy change for the target reaction by summing the enthalpy changes of the reactions used:
Δ H 0 = Δ H 1 0 + 2Δ H 3 0 + Δ H − 4 0
Δ H 0 = − 110.5 + 2 ( − 285.5 ) + 726.6 kJ/mol
Δ H 0 = − 110.5 − 571 + 726.6 kJ/mol
Δ H 0 = 45.1 kJ/mol − 571 kJ/mol
Δ H 0 = − 661.5 + 726.6 = 45.1 kJ/mol
Final Answer Therefore, the enthalpy change for the reaction is 45.1 kJ/mol .
Examples
This calculation is crucial in chemical engineering for designing reactors and optimizing reaction conditions. For instance, if we want to produce methanol from carbon monoxide and hydrogen, knowing the enthalpy change helps determine the amount of heat that needs to be supplied or removed to maintain the reaction at a desired temperature. This ensures efficient and safe operation of the chemical plant, reducing energy consumption and preventing thermal runaways.
The enthalpy change for the reaction C O ( g ) + 2 H 2 ( g ) → C H 3 O H ( l ) is calculated to be 45.1 kJ/mol using Hess's Law. By manipulating the given reactions and applying their enthalpy changes, we derived the desired equation. This process involves reversing and combining the initial reactions systematically.
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