Consider the following intermediate chemical equations:
[tex]
\begin{array}{ll}
P_4(s)+6 Cl_2(g) \rightarrow 4 PCl_3(g) & \Delta H_1=-2,439 kJ \\
4 PCl_5(g) \rightarrow P_4(s)+10 Cl_2(g) & \Delta H_2=3,438 kJ
\end{array}
[/tex]
What is the enthalpy of the overall chemical reaction [tex]PCl _5(g) \rightarrow PCl _3(g)+ Cl _2(g)[/tex]?
A. -999 kJ
B. -250 kJ
C. 250 kJ
D. 999 kJ
Answer (2)
Divide the first equation by 4: P 4 ( s ) /4 + ( 6/4 ) C l 2 ( g ) → PC l 3 ( g ) with Δ H 1 ′ = − 2439/4 k J .
Reverse the second equation and divide by 4: P 4 ( s ) /4 + ( 10/4 ) C l 2 ( g ) → PC l 5 ( g ) with Δ H 2 ′ = − 3438/4 k J .
Subtract the second modified equation from the first: PC l 5 ( g ) → PC l 3 ( g ) + C l 2 ( g ) .
Calculate the enthalpy change: Δ H = Δ H 1 ′ − Δ H 2 ′ = ( − 2439 + 3438 ) /4 = 249.75 k J . The final answer is 250. k J .
Explanation
Problem Analysis We are given two chemical equations:
P 4 ( s ) + 6 C l 2 ( g ) → 4 PC l 3 ( g ) with Δ H 1 = − 2 , 439 k J
4 PC l 5 ( g ) → P 4 ( s ) + 10 C l 2 ( g ) with Δ H 2 = 3 , 438 k J
Our target reaction is: PC l 5 ( g ) → PC l 3 ( g ) + C l 2 ( g )
We need to find the enthalpy change for the target reaction.
Equation Manipulation To find the enthalpy change for the target reaction, we will manipulate the given equations to match the target reaction and then apply Hess's Law. Hess's Law states that the enthalpy change of an overall reaction is the sum of the enthalpy changes of the individual reactions.
First, we reverse the second equation and divide it by 4:
P_4(s)/4 + (10/4)Cl_2(g) \rightarrow PCl_5(g)"> − [ 4 PC l 5 ( g ) → P 4 ( s ) + 10 C l 2 ( g )] /4 => P 4 ( s ) /4 + ( 10/4 ) C l 2 ( g ) → PC l 5 ( g ) with Δ H 2 ′ = − 3438/4 k J
Next, we divide the first equation by 4:
P_4(s)/4 + (6/4)Cl_2(g) \rightarrow PCl_3(g)"> [ P 4 ( s ) + 6 C l 2 ( g ) → 4 PC l 3 ( g )] /4 => P 4 ( s ) /4 + ( 6/4 ) C l 2 ( g ) → PC l 3 ( g ) with Δ H 1 ′ = − 2439/4 k J
Adding Modified Equations Now, we add the two modified equations:
P 4 ( s ) /4 + ( 10/4 ) C l 2 ( g ) → PC l 5 ( g ) + P 4 ( s ) /4 + ( 6/4 ) C l 2 ( g ) → PC l 3 ( g )
This gives us:
P 4 ( s ) /4 + ( 10/4 ) C l 2 ( g ) + P 4 ( s ) /4 + ( 6/4 ) C l 2 ( g ) → PC l 5 ( g ) + PC l 3 ( g )
Enthalpy Calculation However, we are interested in the reaction PC l 5 ( g ) → PC l 3 ( g ) + C l 2 ( g ) .
So, we need to calculate the enthalpy change for the overall reaction by adding the enthalpy changes of the modified reactions:
Δ H = Δ H 1 ′ + Δ H 2 ′ = − 2439/4 + ( − 3438/4 )
Δ H = ( − 2439 − 3438 ) /4 = − 5877/4 = − 1469.25
Corrected Enthalpy Calculation The enthalpy change for the reaction PC l 5 ( g ) → PC l 3 ( g ) + C l 2 ( g ) is -1469.25 kJ. However, this value is not among the options provided. Let's re-examine the problem statement and the given equations.
The target reaction can be obtained by reversing the first reaction, dividing by 4, and reversing the second reaction and dividing by 4, then adding them. So:
Δ H = Δ H 1 ′ + Δ H 2 ′ = − 2439/4 + ( − 3438/4 ) = ( − 2439 − 3438 ) /4 = − 5877/4 = − 1469.25 k J
However, we made an error in the initial setup. We should reverse the second equation and divide by 4, which gives us:
P 4 ( s ) /4 + ( 10/4 ) C l 2 ( g ) → PC l 5 ( g ) with Δ H 2 ′ = − 3438/4 k J
Then, we divide the first equation by 4:
P 4 ( s ) /4 + ( 6/4 ) C l 2 ( g ) → PC l 3 ( g ) with Δ H 1 ′ = − 2439/4 k J
Subtracting the first modified equation from the second modified equation:
[ P 4 ( s ) /4 + ( 6/4 ) C l 2 ( g ) → PC l 3 ( g )] − [ P 4 ( s ) /4 + ( 10/4 ) C l 2 ( g ) → PC l 5 ( g )]
PC l 5 ( g ) → PC l 3 ( g ) + ( 10/4 − 6/4 ) C l 2 ( g )
PC l 5 ( g ) → PC l 3 ( g ) + C l 2 ( g )
Δ H = Δ H 1 ′ − Δ H 2 ′ = − 2439/4 − ( − 3438/4 ) = ( − 2439 + 3438 ) /4 = 999/4 = 249.75 k J
Rounding to the nearest whole number, we get 250 kJ.
Final Answer The enthalpy of the overall chemical reaction PC l 5 ( g ) → PC l 3 ( g ) + C l 2 ( g ) is approximately 250 kJ.
Examples
Hess's Law is a fundamental concept in chemistry that allows us to calculate the enthalpy change of a reaction by using the enthalpy changes of other reactions. For example, if you want to find the energy released or absorbed during the formation of a complex molecule from simpler substances, you can break down the process into smaller, well-known steps. By knowing the enthalpy changes of these individual steps, you can add them up to find the overall enthalpy change for the entire process. This is useful in designing chemical reactions, understanding energy requirements, and optimizing industrial processes to be more efficient and cost-effective. For instance, in the production of fertilizers, understanding the enthalpy changes helps in controlling the reaction conditions to maximize yield and minimize energy consumption.
The enthalpy change for the reaction PC l 5 ( g ) → PC l 3 ( g ) + C l 2 ( g ) is approximately 250 kJ, calculated using Hess's Law by manipulating given chemical equations. This involved reversing one equation, modifying them correctly, and then combining their enthalpy changes. Thus, the correct answer is Option C: 250 kJ.
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