Answer :
To determine the value of \(\Delta H\) for the desired reaction \(X + 2Y \rightarrow 2Z\), 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 for any series of steps that leads to the final reaction.
Given the reactions and their enthalpy changes are:
1. \(W + X \rightarrow 2Y \quad (\Delta H = -200 \text{ kcal})\)
2. \(2W + 3X \rightarrow 2Z + 2Y \quad (\Delta H = -150 \text{ kcal})\)
Step-by-Step Solution:
1. Write down the desired reaction:
[tex]\[ X + 2Y \rightarrow 2Z \][/tex]
2. Analyze and modify the given reactions to facilitate combining them to get the desired reaction:
- The first reaction is \(W + X \rightarrow 2Y\). This reaction gives us 2Y on the right, which is good, but it introduces W and an additional X that we don't need directly in the final reaction.
- The second reaction is \(2W + 3X \rightarrow 2Z + 2Y\). This reaction provides the desired amount of 2Z on the right but also introduces 2Y and additional terms involving W and X.
3. Reverse the second reaction to manipulate it:
- When a reaction is reversed, the sign of \(\Delta H\) also changes. Thus, reversing the second equation:
[tex]\[ 2Z + 2Y \rightarrow 2W + 3X \quad (\Delta H = +150 \text{ kcal}) \][/tex]
4. Combine the two reactions:
- Add the first reaction and the reversed second reaction together to cancel out unwanted terms:
[tex]\[ (W + X \rightarrow 2Y) \][/tex]
[tex]\[ + (2Z + 2Y \rightarrow 2W + 3X) \][/tex]
Combine them as:
[tex]\[ W + X + 2Z + 2Y \rightarrow 2Y + 2W + 3X \][/tex]
5. Cancel out common terms on both sides:
[tex]\[ 2Y \text{ on the left cancels with } 2Y \text{ on the right} \][/tex]
[tex]\[ W \text{ on the left cancels out with one } W \text{ on the right, leaving 1 } W \text{ on the right} \][/tex]
[tex]\[ X \text{ on the left cancels with 1 } X \text{ on the right, leaving 2 } X \text{ on the right} \][/tex]
This simplifies to:
[tex]\[ 2Z + W + X \rightarrow 3X\][/tex]
To match the desired \(X + 2Y \rightarrow 2Z\) and cancel out the remaining terms, we see that we need:
[tex]\[ X + 2Y \rightarrow 2Z\][/tex]
6. Sum the enthalpy changes of individual steps:
- For direct addition:
[tex]\[ \Delta H = -200 \text{ kcal} + 150 \text{ kcal} \][/tex]
[tex]\[ \Delta H = -350 \text{ kcal} \][/tex]
Therefore, the value of \(\Delta H\) for the reaction \(X + 2Y \rightarrow 2Z\) is \(-350\) kcal.
The correct choice is:
[tex]\[ \boxed{-350 \text{ kcal}} \][/tex]
Given the reactions and their enthalpy changes are:
1. \(W + X \rightarrow 2Y \quad (\Delta H = -200 \text{ kcal})\)
2. \(2W + 3X \rightarrow 2Z + 2Y \quad (\Delta H = -150 \text{ kcal})\)
Step-by-Step Solution:
1. Write down the desired reaction:
[tex]\[ X + 2Y \rightarrow 2Z \][/tex]
2. Analyze and modify the given reactions to facilitate combining them to get the desired reaction:
- The first reaction is \(W + X \rightarrow 2Y\). This reaction gives us 2Y on the right, which is good, but it introduces W and an additional X that we don't need directly in the final reaction.
- The second reaction is \(2W + 3X \rightarrow 2Z + 2Y\). This reaction provides the desired amount of 2Z on the right but also introduces 2Y and additional terms involving W and X.
3. Reverse the second reaction to manipulate it:
- When a reaction is reversed, the sign of \(\Delta H\) also changes. Thus, reversing the second equation:
[tex]\[ 2Z + 2Y \rightarrow 2W + 3X \quad (\Delta H = +150 \text{ kcal}) \][/tex]
4. Combine the two reactions:
- Add the first reaction and the reversed second reaction together to cancel out unwanted terms:
[tex]\[ (W + X \rightarrow 2Y) \][/tex]
[tex]\[ + (2Z + 2Y \rightarrow 2W + 3X) \][/tex]
Combine them as:
[tex]\[ W + X + 2Z + 2Y \rightarrow 2Y + 2W + 3X \][/tex]
5. Cancel out common terms on both sides:
[tex]\[ 2Y \text{ on the left cancels with } 2Y \text{ on the right} \][/tex]
[tex]\[ W \text{ on the left cancels out with one } W \text{ on the right, leaving 1 } W \text{ on the right} \][/tex]
[tex]\[ X \text{ on the left cancels with 1 } X \text{ on the right, leaving 2 } X \text{ on the right} \][/tex]
This simplifies to:
[tex]\[ 2Z + W + X \rightarrow 3X\][/tex]
To match the desired \(X + 2Y \rightarrow 2Z\) and cancel out the remaining terms, we see that we need:
[tex]\[ X + 2Y \rightarrow 2Z\][/tex]
6. Sum the enthalpy changes of individual steps:
- For direct addition:
[tex]\[ \Delta H = -200 \text{ kcal} + 150 \text{ kcal} \][/tex]
[tex]\[ \Delta H = -350 \text{ kcal} \][/tex]
Therefore, the value of \(\Delta H\) for the reaction \(X + 2Y \rightarrow 2Z\) is \(-350\) kcal.
The correct choice is:
[tex]\[ \boxed{-350 \text{ kcal}} \][/tex]
