etylene, c₂h₄, reacts with hydrogen to produce methane, ch₄.
c₂h₄ (g) + 2 h₂ (g) → 2 ch₄ (g)
using hesss law, calculate the enthalpy change, δh⁰, for the above reaction:
2 c₂h₆ (g) + 7 o₂ (g) → 4 co₂ (g) + 6 h₂o (l); δh⁰ = -3120.8 kj
ch₄ (g) + 2 o₂ (g) → co₂ (g) + 2 h₂o (l); δh⁰ = -890.3 kj
c₂h₆ (g) → c₂h₄ (g) + h₂ (g); δh⁰ = +136.3 kj
h₂ (g) + 1/2 o₂ (g);→ h₂o (l); δh⁰ = -285.8 kj
report your answer in kilojoules to the correct number of sig figs, but do not include units in your answer.

question 4
0.5 pts
using bond dissociation energies, determine the approximate enthalpy of reaction for the following reaction:
2 h₂s (g) + 3 o₂ (g) → 2 so₂ (g) + 2h₂o (g)
note that bond dissociation energies can be found in the tables of chemical data posted on canvas. report the answer in kilojoules to 3 sig figs, but do not include units in your answer.

Question

etylene, c₂h₄, reacts with hydrogen to produce methane, ch₄.
c₂h₄ (g) + 2 h₂ (g) → 2 ch₄ (g)
using hesss law, calculate the enthalpy change, δh⁰, for the above reaction:
2 c₂h₆ (g) + 7 o₂ (g) → 4 co₂ (g) + 6 h₂o (l); δh⁰ = -3120.8 kj
ch₄ (g) + 2 o₂ (g) → co₂ (g) + 2 h₂o (l); δh⁰ = -890.3 kj
c₂h₆ (g) → c₂h₄ (g) + h₂ (g); δh⁰ = +136.3 kj
h₂ (g) + 1/2 o₂ (g);→ h₂o (l); δh⁰ = -285.8 kj
report your answer in kilojoules to the correct number of sig figs, but do not include units in your answer.

question 4
0.5 pts
using bond dissociation energies, determine the approximate enthalpy of reaction for the following reaction:
2 h₂s (g) + 3 o₂ (g) → 2 so₂ (g) + 2h₂o (g)
note that bond dissociation energies can be found in the tables of chemical data posted on canvas. report the answer in kilojoules to 3 sig figs, but do not include units in your answer.

Accepted Answer

### First Sub - Question (Using Hess's Law)
# Explanation:
## Step 1: List the target reaction and given reactions
Target reaction: $\ce{C2H4 (g) + 2 H2 (g) -> 2 CH4 (g)}$
Given reactions:
1. $\ce{2 C2H6 (g) + 7 O2 (g) -> 4 CO2 (g) + 6 H2O (l)}$; $\Delta H^{\circ}_1=- 3120.8\space kJ$
2. $\ce{CH4 (g) + 2 O2 (g) -> CO2 (g) + 2 H2O (l)}$; $\Delta H^{\circ}_2 = - 890.3\space kJ$
3. $\ce{C2H6 (g) -> C2H4 (g) + H2 (g)}$; $\Delta H^{\circ}_3=+136.3\space kJ$
4. $\ce{H2 (g) + 1/2 O2 (g)-> H2O (l)}$; $\Delta H^{\circ}_4=-285.8\space kJ$

## Step 2: Manipulate the given reactions to match the target reaction
- Reverse reaction 3: $\ce{C2H4 (g) + H2 (g)-> C2H6 (g)}$; $\Delta H^{\circ}_{3r}=-\Delta H^{\circ}_3=- 136.3\space kJ$
- Multiply reaction 2 by 2: $\ce{2CH4 (g) + 4 O2 (g) -> 2CO2 (g) + 4 H2O (l)}$; $\Delta H^{\circ}_{2m}=2	imes\Delta H^{\circ}_2=2	imes(- 890.3)=-1780.6\space kJ$ (reverse this reaction for later: $\ce{2CO2 (g) + 4 H2O (l)-> 2CH4 (g) + 4 O2 (g)}$; $\Delta H^{\circ}_{2mr}=- \Delta H^{\circ}_{2m}=1780.6\space kJ$)
- Divide reaction 1 by 2: $\ce{C2H6 (g) + 3.5 O2 (g) -> 2 CO2 (g) + 3 H2O (l)}$; $\Delta H^{\circ}_{1d}=\frac{\Delta H^{\circ}_1}{2}=\frac{-3120.8}{2}=- 1560.4\space kJ$
- Multiply reaction 4 by 2: $\ce{2H2 (g) + O2 (g)-> 2H2O (l)}$; $\Delta H^{\circ}_{4m}=2	imes\Delta H^{\circ}_4=2	imes(-285.8)=-571.6\space kJ$

Now, let's combine the reactions:
Start with reversed reaction 3: $\ce{C2H4 (g) + H2 (g)-> C2H6 (g)}$; $\Delta H=-136.3$
Add reaction 1d: $\ce{C2H6 (g) + 3.5 O2 (g) -> 2 CO2 (g) + 3 H2O (l)}$; $\Delta H=-1560.4$
Add reversed reaction 2m: $\ce{2CO2 (g) + 4 H2O (l)-> 2CH4 (g) + 4 O2 (g)}$; $\Delta H = 1780.6$
Add reaction 4m: $\ce{2H2 (g) + O2 (g)-> 2H2O (l)}$; $\Delta H=-571.6$

Now, sum up the left - hand side and right - hand side species:
Left: $\ce{C2H4 (g) + H2 (g)+C2H6 (g) + 3.5 O2 (g)+2CO2 (g) + 4 H2O (l)+2H2 (g) + O2 (g)}$
Right: $\ce{C2H6 (g) + 2 CO2 (g) + 3 H2O (l)+2CH4 (g) + 4 O2 (g)+2H2O (l)}$

After canceling out common species:
$\ce{C2H4 (g) + 3H2 (g)-> 2CH4 (g) + H2O (l)}$ Wait, no, let's re - do the manipulation.

Alternative approach:
We want to get $\ce{C2H4 + 2H2->2CH4}$

From reaction 3: $\ce{C2H6 = C2H4 + H2}$ (reverse: $\ce{C2H4 + H2 = C2H6}$; $\Delta H=-136.3$)

From reaction 2: $\ce{CH4 + 2O2=CO2 + 2H2O}$ (multiply by 2 and reverse: $\ce{2CO2 + 4H2O = 2CH4 + 4O2}$; $\Delta H = 2	imes890.3 = 1780.6$)

From reaction 1: $\ce{2C2H6 + 7O2=4CO2 + 6H2O}$ (divide by 2: $\ce{C2H6 + 3.5O2=2CO2 + 3H2O}$; $\Delta H=-3120.8/2=-1560.4$)

From reaction 4: $\ce{H2 + 0.5O2=H2O}$ (multiply by 2: $\ce{2H2 + O2=2H2O}$; $\Delta H = 2	imes(-285.8)=-571.6$)

Now, add the reversed reaction 3, reaction 1 (divided by 2), reversed reaction 2 (multiplied by 2), and reaction 4 (multiplied by 2):

Reversed reaction 3: $\ce{C2H4 + H2 = C2H6}$; $\Delta H=-136.3$

Reaction 1 (divided by 2): $\ce{C2H6 + 3.5O2=2CO2 + 3H2O}$; $\Delta H=-1560.4$

Reversed reaction 2 (multiplied by 2): $\ce{2CO2 + 4H2O=2CH4 + 4O2}$; $\Delta H = 1780.6$

Reaction 4 (multiplied by 2): $\ce{2H2 + O2=2H2O}$; $\Delta H=-571.6$

Now, sum the $\Delta H$ values:
$\Delta H=-136.3-1560.4 + 1780.6-571.6$
First, $-136.3-1560.4=-1696.7$
Then, $-1696.7 + 1780.6 = 83.9$
Then, $83.9-571.6=-487.7$ Wait, this is wrong. Let's try another way.

Let's express the target reaction in terms of given reactions:

Target reaction: $\ce{C2H4 + 2H2->2CH4}$

We can write:

From reaction 3: $\ce{C2H6 = C2H4 + H2}$ (so $\ce{C2H4 = C2H6 - H2}$)

From reaction 2: $\ce{CH4=\frac{1}{2}(CO2 + 2H2O - 2O2)}$ (multiply by 2: $\ce{2CH4=CO2 + 2H2O - 2O2}$)

From reaction 1: $\ce{2C2H6=4CO2 + 6H2O - 7O2}$ (so $\ce{C2H6 = 2CO2 + 3H2O - 3.5O2}$)

From reaction 4: $\ce{H2=H2O - 0.5O2}$ (multiply by 2: $\ce{2H2=2H2O - O2}$)

Substitute $\ce{C2H4 = C2H6 - H2}$ into target reaction:

$\ce{(C2H6 - H2)+2H2->2CH4}$

$\ce{C2H6 + H2->2CH4}$

Now substitute $\ce{C2H6 = 2CO2 + 3H2O - 3.5O2}$ and $\ce{2CH4=CO2 + 2H2O - 2O2}$ and $\ce{H2=H2O - 0.5O2}$:

$\ce{(2CO2 + 3H2O - 3.5O2)+(H2O - 0.5O2)->(CO2 + 2H2O - 2O2)}$

Left: $2CO2 + 4H2O - 4O2$

Right: $CO2 + 2H2O - 2O2$

Subtract right from left: $CO2 + 2H2O - 2O2 = 0$? No, this is not helpful.

Let's use the method of linear combination.

Let the target reaction be $r = a	imes r1 + b	imes r2 + c	imes r3 + d	imes r4$

Where $r$ is the target reaction, $r1,r2,r3,r4$ are given reactions.

For $\ce{C2H4}$: coefficient in $r$ is 1, in $r3$ is - 1 (since $r3$ is $\ce{C2H6->C2H4 + H2}$, so $\ce{C2H4}$ has coefficient - 1 in reverse of $r3$), so $-c = 1\Rightarrow c=-1$

For $\ce{H2}$: coefficient in $r$ is 2, in $r3$ is - 1 (reverse of $r3$: $\ce{C2H4 + H2->C2H6}$, so $H2$ has coefficient 1), in $r4$ is 1 (per reaction 4), so $1	imes c + 2	imes d=2$. Since $c = - 1$, then $-1+2d = 2\Rightarrow 2d=3\Rightarrow d = 1.5$

For $\ce{CH4}$: coefficient in $r$ is 2, in $r2$ is - 1 (reverse of $r2$: $\ce{CO2 + 2H2O->CH4 + 2O2}$, so $\ce{CH4}$ has coefficient - 1), so $-2b=2\Rightarrow b=-2$

For $\ce{C2H6}$: coefficient in $r$ is 0, in $r1$ is 1, in $r3$ is 1 (reverse of $r3$: $\ce{C2H4 + H2->C2H6}$, so $\ce{C2H6}$ has coefficient 1), so $0.5	imes a+1	imes c=0$. Since $c=-1$, then $0.5a-1 = 0\Rightarrow a = 2$

Now check $\ce{O2}$:

In $r1$: $a = 2$, so $\ce{O2}$ coefficient is $3.5	imes a=7$

In $r2$: $b=-2$, so $\ce{O2}$ coefficient is $-2	imes2=-4$

In $r3$: $c=-1$, so $\ce{O2}$ coefficient is 0

In $r4$: $d = 1.5$, so $\ce{O2}$ coefficient is $-0.5	imes1.5=-0.75$

Total $\ce{O2}$ coefficient: $7-4 - 0.75 = 2.25$? No, this is wrong.

Let's try a better way:

Target reaction: $\ce{C2H4 + 2H2->2CH4}$

We can get $\ce{C2H4}$ from reverse of $r3$: $\ce{C2H4 + H2->C2H6}$; $\Delta H=-136.3$

We can get $\ce{2CH4}$ from reverse of 2 times $r2$: $\ce{2CO2 + 4H2O->2CH4 + 4O2}$; $\Delta H=2	imes890.3 = 1780.6$

We can get $\ce{C2H6}$ consumption from 0.5 times $r1$: $\ce{C2H6 + 3.5O2->2CO2 + 3H2O}$; $\Delta H=-3120.8/2=-1560.4$

We can get $\ce{2H2}$ from 2 times $r4$: $\ce{2H2 + O2->2H2O}$; $\Delta H=2	imes(-285.8)=-571.6$

Now add all these reactions:

$\ce{C2H4 + H2->C2H6}$; $\Delta H=-136.3$

$\ce{C2H6 + 3.5O2->2CO2 + 3H2O}$; $\Delta H=-1560.4$

$\ce{2CO2 + 4H2O->2CH4 + 4O2}$; $\Delta H=1780.6$

$\ce{2H2 + O2->2H2O}$; $\Delta H=-571.6$

Now, sum the equations:

Left: $\ce{C2H4 + H2 + C2H6 + 3.5O2 + 2CO2 + 4H2O + 2H2 + O2}$

Right: $\ce{C2H6 + 2CO2 + 3H2O + 2CH4 + 4O2 + 2H2O}$

Cancel out $\ce{C2H6}$, $\ce{2CO2}$:

Left: $\ce{C2H4 + 3H2 + 4.5O2 + 4H2O}$

Right: $\ce{5H2O + 2CH4 + 4O2}$

Subtract right from left: $\ce{C2H4 + 3H2 + 4.5O2 + 4H2O-5H2O - 2CH4 - 4O2}=\ce{C2H4 + 3H2 + 0.5O2 - H2O - 2CH4}=0$? No, this is incorrect. I must have made a mistake in the combination.

Let's start over.

We know that:

Reaction 3: $\ce{C2H6 -> C2H4 + H2}$ $\Delta H = + 136.3$ (so $\ce{C2H4 = C2H6 - H2}$)

Reaction 2: $\ce{CH4 + 2O2->CO2 + 2H2O}$ $\Delta H=-890.3$ (multiply by 2: $\ce{2CH4 + 4O2->2CO2 + 4H2O}$ $\Delta H=-1780.6$; reverse: $\ce{2CO2 + 4H2O->2CH4 + 4O2}$ $\Delta H = + 1780.6$)

Reaction 1: $\ce{2C2H6 + 7O2->4CO2 + 6H2O}$ $\Delta H=-3120.8$ (divide by 2: $\ce{C2H6 + 3.5O2->2CO2 + 3H2O}$ $\Delta H=-1560.4$)

Reaction 4: $\ce{H2 + 0.5O2->H2O}$ $\Delta H=-285.8$ (multiply by 2: $\ce{2H2 + O2->2H2O}$ $\Delta H=-571.6$)

We want $\ce{C2H4 + 2H2->2CH4}$

Let's express the target reaction as:

$\ce{C2H4 + 2H2->2CH4}$

Substitute $\ce{C2H4}$ from reaction 3: $\ce{(C2H6 - H2)+2H2->2CH4}$ $\Rightarrow\ce{C2H6 + H2->2CH4}$

Now, we can get $\ce{C2H6 + H2->2CH4}$ from other reactions.

We have:

From reaction 1 (divided by 2): $\ce{C2H6 + 3.5O2->2CO2 + 3H2O}$ $\Delta H=-1560.4$

From reaction 2 (reversed and multiplied by 2): $\ce{2CO2 + 4H2O->2CH4 + 4O2}$ $\Delta H = + 1780.6$

From reaction 4 (multiplied by 1): $\ce{H2 + 0.5O2->H2O}$ $\Delta H=-285.8$ (we need $\ce{H2}$ in the reactant, so this is correct)

Now add reaction 1 (divided by 2), reaction 2 (reversed and multiplied by 2), and reaction

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

First Sub - Question (Using Hess's Law)

Step 1: List the target reaction and given reactions

Step 2: Manipulate the given reactions to match the target reaction

Answer:

Step 1: List the target reaction and given reactions

Step 2: Manipulate the given reactions to match the target reaction