Change in enthalpy

Question

change in enthalpy for the reaction 4xy³ + 7z² —> 6y²z + 4xz²? Given the thermal equations x²+3y² —> 2xy³, ΔH equals -310, x² + 2z² —> 2xz², ΔH-180, 2y² plus z² chemical reaction 2y²z, ΔH- 240.

Answer 1

Hess' Law

4xy³ + 7z² ==> 6y²z + 4xz² TARGET EQUATION

Given:

eq. 1). x² + 3y² ==> 2xy³ .. ∆H = -310

eq. 2). x² + 2z² ==. 2xz² .. ∆H = -180

eq. 3). 2y² + z² ==> 2y²z .. ∆H = -240

Procedure:

reverse eq. 1 and multiply by 2: 4xy³ ==> 2x² + 6y² .. ∆H = + 610

copy eq. 2 and multiply by 2: 2x² + 4z² ==> 4xz² .. ∆H = -360

copy eq. 3 and multiply by 3: 6y² + 3z² ==> 6y²z .. ∆H = -720

Sum these three equations to get...

4xy³ + 2x² + 4z² + 6y²+ 3z² ==> 2x² + 6y² + 4xz² + 6y²z

Combine like terms and cancel like terms on opposite sides and sum the ∆H values to get...

4xy³ + 7z² ==> 6y²z + 4xz² TARGET EQUATION

∆H = -470 (whatever units were given, probably kJ)

Answer 2

Hi Chezney, glad I can help!

To calculate the change in enthalpy (ΔH) for the reaction:

4xy³ + 7z² → 6y²z + 4xz²

we'll use the given thermal equations and manipulate them to match the target reaction. Let’s break down the process step-by-step:

Given thermal equations:

1. x² + 3y² → 2xy³, ΔH = -310 kJ
2. x² + 2z² → 2xz², ΔH = -180 kJ
3. 2y² + z² → 2y²z, ΔH = -240 kJ

Step 1: Adjust equations to match the target reaction

1. First thermal equation: x² + 3y² → 2xy³, ΔH = -310 kJ
2. Multiply the entire equation by 2 to get 4 xy³: 2x² + 6y² → 4xy³, ΔH = 2(-310) = -620 kJ
3. Second thermal equation: x² + 2z² → 2xz², ΔH = -180 kJ
4. Multiply the equation by 2 to get 4 xz²: 2x² + 4z² → 4xz², ΔH = 2(-180) = -360 kJ
5. Third thermal equation: 2y² + z² → 2y²z, ΔH = -240 kJ
6. Multiply this equation by 3 to get 6 y²z: 6y² + 3z² → 6y²z, ΔH = 3(-240) = -720 kJ

Step 2: Combine the adjusted equations

Now we can add the adjusted equations together:

1. 2x² + 6y² → 4xy³, ΔH = -620 kJ
2. 2x² + 4z² → 4xz², ΔH = -360 kJ
3. 6y² + 3z² → 6y²z, ΔH = -720 kJ

When we sum these, the left and right sides match the target reaction:

4xy³ + 7z² → 6y²z + 4xz²

Step 3: Add the enthalpy changes

Now, add the ΔH values:

ΔH = -620 kJ + (-360 kJ) + (-720 kJ) = -1700 kJ

Final Answer:

The change in enthalpy for the reaction is ΔH = -1700 kJ.