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makkiz [27]
3 years ago
10

For the following reactions, the ΔH° rxn is not equal to ΔH° f for the product except for ________.

Chemistry
1 answer:
kirill [66]3 years ago
5 0

Answer:

The correct option is: e. 2C(s, graphite) + 2H₂(g) → C₂H₄ (g)

Explanation:

The standard enthalpy of the reaction (\Delta H_{r}^{\circ}) is the change in enthalpy associated with a given chemical reaction. It can be calculated by the standard enthalpies of formation of the products and reactants.

\Delta H_{r}^{\circ} = \sum \Delta H_{f}^{\circ} (Products) - \sum \Delta H_{f}^{\circ} (reactants)

The standard enthalpy of formation (\Delta H_{f}^{\circ}) <u><em>of an elements that is present in its standard state is zero.</em></u>

Therefore,

a. H₂O (l) + 1/2 O₂ (g) → H₂O₂(l)

\Delta H_{r}^{\circ} = \left [\Delta H_{f}^{\circ} (H_{2}O_{2}, l) \right ] - \left [\Delta H_{f}^{\circ} (H_{2}O, l)+ 1/2\times \Delta H_{f}^{\circ} (O_{2}, g)\right]

Here, the \Delta H_{f}^{\circ} (O_{2}, g) = 0 kJ/mol

→ \Delta H_{r}^{\circ} = \left [\Delta H_{f}^{\circ} (H_{2}O_{2}, l) \right ] - \left [\Delta H_{f}^{\circ} (H_{2}O, l)\right] \neq \Delta H_{f}^{\circ} (Products)

b. N₂ (g) + O₂ (g) → 2NO (g)

\Delta H_{r}^{\circ} = \left [2\times \Delta H_{f}^{\circ} (NO, g) \right ] - \left [\Delta H_{f}^{\circ} (N_{2}, g)+\Delta H_{f}^{\circ} (O_{2}, g)\right]

Here, the \Delta H_{f}^{\circ} (O_{2}, g) = \Delta H_{f}^{\circ} (N_{2}, g)= 0 kJ/mol

→ \Delta H_{r}^{\circ} = \left [2\times \Delta H_{f}^{\circ} (NO, g) \right ] \neq \Delta H_{f}^{\circ} (Product, NO)

c. 2H₂ (g) + O₂ (g) → 2H₂O (g)

\Delta H_{r}^{\circ} = \left [2\times \Delta H_{f}^{\circ} (H_{2}O, g) \right ] - \left [2\times \Delta H_{f}^{\circ} (H_{2}, g)+\Delta H_{f}^{\circ} (O_{2}, g)\right]

Here, the \Delta H_{f}^{\circ} (O_{2}, g) = \Delta H_{f}^{\circ} (H_{2}, g)= 0 kJ/mol

→ \Delta H_{r}^{\circ} = \left [2\times \Delta H_{f}^{\circ} (H_{2}O, g) \right ] \neq \Delta H_{f}^{\circ} (Product, H_{2}O)

d. 2H₂ (g) + O₂ (g) → 2H₂O (l)

\Delta H_{r}^{\circ} = \left [2\times \Delta H_{f}^{\circ} (H_{2}O, l) \right ] - \left [2\times \Delta H_{f}^{\circ} (H_{2}, g)+\Delta H_{f}^{\circ} (O_{2}, g)\right]

Here, the \Delta H_{f}^{\circ} (O_{2}, g) = \Delta H_{f}^{\circ} (H_{2}, g)= 0 kJ/mol

→ \Delta H_{r}^{\circ} = \left [2\times \Delta H_{f}^{\circ} (H_{2}O, l) \right ] \neq \Delta H_{f}^{\circ} (Product, H_{2}O)

e. 2C(s, graphite) + 2H₂(g) → C₂H₄ (g)

\Delta H_{r}^{\circ} = \left [\Delta H_{f}^{\circ} (C_{2}H_{4}, g) \right ] - \left [2\times \Delta H_{f}^{\circ} (C, s graphite)+2\times \Delta H_{f}^{\circ} (H_{2}, g)\right]

Here, the \Delta H_{f}^{\circ} (C, s, graphite) = \Delta H_{f}^{\circ} (H_{2}, g)= 0 kJ/mol

→ \Delta H_{r}^{\circ} = \left [\Delta H_{f}^{\circ} (C_{2}H_{4}, g) \right ] = \Delta H_{f}^{\circ} (Product, C_{2}H_{4})

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Answer:

C. 1, 4–dimethyl–1–cyclohexene

D. 4, 6–dichloro–1–heptene

Explanation:

C. Determination of the name of the compound.

To name the compound given above we must observe the following:

1. Determine the functional group of the compound. In this case, the functional group is the double bond (–C=C–).

2. Determine the parent name by counting the number of carbon that makes up the ring. In this case, 6 carbon makes up the ring. Hence the parent name is cyclohexene.

3. Identify the substituent group attached to the compound. In this case, two methyl group i.e –CH₃ are attached.

4. Locate the position of the substituent group attached by giving the functional group the lowest possible count. In this case, the functional group is at carbon 1, the first methyl group is at carbon 1 and the 2nd methyl group is at carbon 4.

5. Combine the above to obtain the name. Thus, the name of the compound is:

1, 4–dimethyl–1–cyclohexene.

D. Determination of the name of the compound.

To name the compound given above we must observe the following:

1. Determine the functional group of the compound. In this case, the functional group is the double bond (–C=C–).

2. Determine the parent name by counting the number of carbon that makes up the chain. In this case, 7 carbons makes up the chain. Hence, the parent name of the compound is heptene.

3. Identify the substituent group attached to the compound. In this case, two chlorine (Chloro) i.e –Cl are attached.

4. Locate the position of the substituent group attached by giving the functional group the lowest possible count. In this case, the functional group is at carbon 1, the first chlorine is at carbon 4 and the 2nd is at carbon 6.

5. Combine the above to obtain the name. Thus, the name of the compound is:

4, 6–dichloro–1–heptene

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