Question

question 6
given the following two reactions:
2a + b ⇌ 3c δhrxn = -26 kj
2c + d ⇌ 2b δhrxn = -44 kj
calculate δhrxn for the following reaction (in kj).
4c ⇌ 4a + d δhrxn =???

Explanation:

Step1: Label the reactions

Let Reaction 1: \( 2A + B ightleftharpoons 3C \), \( \Delta H_{rxn1} = -26 \, \text{kJ} \)
Reaction 2: \( 2C + D ightleftharpoons 2B \), \( \Delta H_{rxn2} = -44 \, \text{kJ} \)
Target Reaction: \( 4C ightleftharpoons 4A + D \)

Step2: Manipulate Reaction 1

Reverse Reaction 1 and multiply by 2:
\( 6C ightleftharpoons 4A + 2B \), \( \Delta H_{rxn1'} = -2 \times (-26) = 52 \, \text{kJ} \)

Step3: Manipulate Reaction 2

Reverse Reaction 2:
\( 2B ightleftharpoons 2C + D \), \( \Delta H_{rxn2'} = -(-44) = 44 \, \text{kJ} \)

Step4: Add the manipulated reactions

Add \( 6C ightleftharpoons 4A + 2B \) (with \( \Delta H = 52 \, \text{kJ} \)) and \( 2B ightleftharpoons 2C + D \) (with \( \Delta H = 44 \, \text{kJ} \)):
\( 6C + 2B ightleftharpoons 4A + 2B + 2C + D \)
Simplify (cancel \( 2B \)): \( 4C ightleftharpoons 4A + D \)

Step5: Calculate \( \Delta H_{rxn} \)

\( \Delta H_{rxn} = \Delta H_{rxn1'} + \Delta H_{rxn2'} = 52 + 44 = 96 \, \text{kJ} \)? Wait, no—wait, let's recheck. Wait, when we reverse Reaction 1, the original is \( 2A + B \to 3C \) (exothermic, \( \Delta H = -26 \)). Reversing gives \( 3C \to 2A + B \), \( \Delta H = +26 \). Then multiplying by 2: \( 6C \to 4A + 2B \), \( \Delta H = 2 \times 26 = 52 \, \text{kJ} \). Reaction 2 reversed: \( 2B \to 2C + D \), \( \Delta H = +44 \, \text{kJ} \) (since original was \( 2C + D \to 2B \), \( \Delta H = -44 \), so reverse is \( +44 \)). Now add the two: \( 6C + 2B \to 4A + 2B + 2C + D \). Cancel \( 2B \): \( 6C \to 4A + 2C + D \). Then subtract \( 2C \) from both sides: \( 4C \to 4A + D \). Wait, my earlier simplification was wrong. Let's do it again.

Wait, Reaction 1' (after reversing and doubling): \( 6C ightarrow 4A + 2B \), \( \Delta H = 52 \, \text{kJ} \)
Reaction 2' (after reversing): \( 2B ightarrow 2C + D \), \( \Delta H = 44 \, \text{kJ} \)

Now add them: \( 6C + 2B ightarrow 4A + 2B + 2C + D \)
Subtract \( 2B \) from both sides: \( 6C ightarrow 4A + 2C + D \)
Subtract \( 2C \) from both sides: \( 4C ightarrow 4A + D \)

Now, \( \Delta H = 52 + 44 = 96 \, \text{kJ} \)? Wait, but let's check with Hess's law again. Wait, maybe I made a sign error. Wait, original Reaction 1: \( 2A + B ightleftharpoons 3C \), \( \Delta H = -26 \) (so forward is exothermic). To get \( 4A + D \) on the right, we need to reverse Reaction 1 (to get \( 3C \to 2A + B \)) and scale, and reverse Reaction 2 (to get \( 2B \to 2C + D \)). Wait, maybe another approach: Let's write the target reaction: \( 4C \to 4A + D \). Let's express this as a combination of Reaction 1 and Reaction 2.

Reaction 1: \( 2A + B \to 3C \), \( \Delta H_1 = -26 \)
Reaction 2: \( 2C + D \to 2B \), \( \Delta H_2 = -44 \)

We need to eliminate \( B \). Let's solve for \( B \) from Reaction 1: \( B = 3C - 2A \) (from \( 2A + B = 3C \)). Substitute into Reaction 2: \( 2C + D \to 2(3C - 2A) \)
\( 2C + D \to 6C - 4A \)
Rearrange: \( 4A + D \to 4C \)
Which is the reverse of the target reaction (\( 4C \to 4A + D \)), so \( \Delta H \) for target is the negative of \( \Delta H \) for \( 4A + D \to 4C \).

Calculate \( \Delta H \) for \( 4A + D \to 4C \):
From Reaction 1: multiply by 2: \( 4A + 2B \to 6C \), \( \Delta H = 2 \times (-26) = -52 \)
From Reaction 2: \( 2C + D \to 2B \), \( \Delta H = -44 \)
Add these two: \( 4A + 2B + 2C + D \to 6C + 2B \)
Simplify: \( 4A + D \to 4C \), \( \Delta H = -52 + (-44) = -96 \, \text{kJ} \)
Therefore, the target reaction \( 4C \to 4A + D \) has \( \Delta H = +96 \, \text{kJ} \)? Wait, no—wait, if \( 4A + D \to 4C \) has \( \Delta H = -96 \), then reversing gives \( 4C \to 4A + D \) with \( \Delta H = +96 \, \text{kJ} \). Wait, but let's check again.

Wait, Reaction 1: \( 2A + B \to 3C \), \( \Delta H = -26 \). Multiply by 2: \( 4A + 2B \to 6C \), \( \Delta H = -52 \) (exothermic, so \( \Delta H \) is negative).
Reaction 2: \( 2C + D \to 2B \), \( \Delta H = -44 \) (exothermic).
Now add these two reactions:
\( 4A + 2B + 2C + D \to 6C + 2B \)
Cancel \( 2B \): \( 4A + D \to 4C \), \( \Delta H = -52 + (-44) = -96 \, \text{kJ} \) (since both are exothermic, adding their \( \Delta H \) gives more exothermic).
Therefore, the reverse reaction \( 4C \to 4A + D \) has \( \Delta H = +96 \, \text{kJ} \) (endothermic). Wait, but earlier when I added the reversed reactions, I think I messed up the signs. Let's redo the Hess's law properly:

Target reaction: \( 4C ightleftharpoons 4A + D \)

We have:
1. \( 2A + B ightleftharpoons 3C \), \( \Delta H_1 = -26 \)
2. \( 2C + D ightleftharpoons 2B \), \( \Delta H_2 = -44 \)

We need to combine 1 and 2 to get \( 4C ightleftharpoons 4A + D \).

Let’s solve for \( B \) from reaction 1: \( B = 3C - 2A \) (from \( 2A + B = 3C \)). Substitute into reaction 2:

\( 2C + D ightleftharpoons 2(3C - 2A) \)
\( 2C + D ightleftharpoons 6C - 4A \)
Rearrange: \( 4A + D ightleftharpoons 4C \)

This is the reverse of the target reaction (\( 4C ightleftharpoons 4A + D \)), so \( \Delta H \) for target is \( -\Delta H \) for \( 4A + D ightleftharpoons 4C \).

Calculate \( \Delta H \) for \( 4A + D ightleftharpoons 4C \):

From reaction 1: multiply by 2: \( 4A + 2B ightleftharpoons 6C \), \( \Delta H = 2 \times (-26) = -52 \)
From reaction 2: \( 2C + D ightleftharpoons 2B \), \( \Delta H = -44 \)

Add these two reactions:
\( 4A + 2B + 2C + D ightleftharpoons 6C + 2B \)
Simplify (cancel \( 2B \)): \( 4A + D ightleftharpoons 4C \)
\( \Delta H = -52 + (-44) = -96 \, \text{kJ} \)

Therefore, the target reaction \( 4C ightleftharpoons 4A + D \) has \( \Delta H = -(-96) = 96 \, \text{kJ} \)? Wait, no—wait, if \( 4A + D ightleftharpoons 4C \) has \( \Delta H = -96 \), then \( 4C ightleftharpoons 4A + D \) is the reverse, so \( \Delta H = +96 \, \text{kJ} \). Wait, but let's check with Hess's law steps again.

Alternative approach:

Target: \( 4C \to 4A + D \)

We can write this as:

- Reverse reaction 1 (multiply by 2): \( 6C \to 4A + 2B \), \( \Delta H = 2 \times 26 = 52 \) (since reversing \( 2A + B \to 3C \) (ΔH=-26) gives \( 3C \to 2A + B \) (ΔH=+26), then ×2: \( 6C \to 4A + 2B \), ΔH=+52)
- Reverse reaction 2: \( 2B \to 2C + D \), \( \Delta H = +44 \) (since original \( 2C + D \to 2B \), ΔH=-44, reverse is +44)

Now add these two reactions:

\( 6C \to 4A + 2B \) (ΔH=52)
\( 2B \to 2C + D \) (ΔH=44)
Sum: \( 6C + 2B \to 4A + 2B + 2C + D \)
Cancel \( 2B \): \( 6C \to 4A + 2C + D \)
Subtract \( 2C \) from both sides: \( 4C \to 4A + D \)

Now, ΔH for this is 52 + 44 = 96 kJ. Yes, that matches. So the enthalpy change for \( 4C ightleftharpoons 4A + D \) is \( +96 \, \text{kJ} \)? Wait, but let's check the signs again. The original reactions:

Reaction 1: \( 2A + B \to 3C \), exothermic (ΔH=-26), so reverse is endothermic (ΔH=+26). Multiply by 2: ΔH=+52.

Reaction 2: \( 2C + D \to 2B \), exothermic (ΔH=-44), so reverse is endothermic (ΔH=+44).

Adding two endothermic reactions: total ΔH is +52 + +44 = +96. So the target reaction (which is the sum) is endothermic, ΔH=+96 kJ. Wait, but the problem says "4C ⇌ 4A + D", so the forward reaction is \( 4C \to 4A + D \), ΔH=+96 kJ.

Wait, but let's verify with another method. Let's assign variables. Let \( \Delta H_f(A) \), \( \Delta H_f(B) \), \( \Delta H_f(C) \), \( \Delta H_f(D) \) be the enthalpies of formation.

For Reaction 1: \( 2\Delta H_f(A) + \Delta H_f(B) - 3\Delta H_f(C) = -26 \) ...(1)

For Reaction 2: \( 2\Delta H_f(C) + \Delta H_f(D) - 2\Delta H_f(B) = -44 \) ...(2)

Target reaction: \( 4\Delta H_f(C) - (4\Delta H_f(A) + \Delta H_f(D)) = \Delta H_{rxn} \)
We need to find \( 4\Delta H_f(C) - 4\Delta H_f(A) - \Delta H_f(D) \).

From (1): \( 2\Delta H_f(A) + \Delta H_f(B) = 3\Delta H_f(C) - 26 \) ...(1a)

From (2): \( 2\Delta H_f(C) + \Delta H_f(D) = 2\Delta H_f(B) - 44 \) ...(2a)

Let's solve (1a) for \( \Delta H_f(B) \): \( \Delta H_f(B) = 3\Delta H_f(C) - 26 - 2\Delta H_f(A) \)

Substitute into (2a):

\( 2\Delta H_f(C) + \Delta H_f(D) = 2(3\Delta H_f(C) - 26 - 2\Delta H_f(A)) - 44 \)
\( 2\Delta H_f(C) + \Delta H_f(D) = 6\Delta H_f(C) - 52 - 4\Delta H_f(A) - 44 \)
\( 2\Delta H_f(C) + \Delta H_f(D) = 6\Delta H_f(C) - 4\Delta H_f(A) - 96 \)
Rearrange:

\( -4\Delta H_f(C) + \Delta H_f(D) + 4\Delta H_f(A) = -96 \)
Multiply both sides by -1:

\( 4\Delta H_f(C) - \Delta H_f(D) - 4\Delta H_f(A) = 96 \)

Which is exactly the target reaction's \( \Delta H_{rxn} \): \( 4\Delta H_f(C) - (4\Delta H_f(A) + \Delta H_f(D)) = 96 \, \text{kJ} \).

Yes, so the enthalpy change is \( +96 \, \text{kJ} \). Wait, but initially I thought maybe a mistake, but the algebra confirms it. So the correct \( \Delta H_{rxn} \) is \( 96 \, \text{kJ} \). Wait, but let's check the sign again. The target reaction is \( 4C \to 4A + D \). If the enthalpy of formation of products (4A + D) is higher than reactants (4C), then the reaction is endothermic, ΔH positive. The calculations show ΔH=+96 kJ.

Wait, but the initial steps: when we reversed Reaction 1 (multiply by 2) and reversed Reaction 2, then added, we got the target reaction with ΔH=52+44=96. The algebraic method also gives 96. So the correct answer is 96 kJ.

Answer:

\( \boldsymbol{96 \, \text{kJ}} \)