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3 years ago

Categorize each hydrocarbon as being saturated or unsaturated.

Chemistry

1 answer:

Bond [772]3 years ago

8 0

Answer :

Saturated hydrocarbon : 3-methyloctane, propane, heptane.

Unsaturated hydrocarbon : cyclopentene, heptyne

Explanation :

Saturated hydrocarbon : These are those hydrocarbons in which all carbon atoms and hydrogen atoms are bonded together with single bonds.

For example : Methane, Ethane, etc.

Unsaturated hydrocarbon : These are those hydrocarbons in which carbon-carbon atom bonded together with double or triple covalent bonds.

For example : Ethene, Butyne, etc.

Therefore, the saturated and unsaturated hydrocarbons are:

Saturated hydrocarbon : 3-methyloctane, propane, heptane.

Unsaturated hydrocarbon : cyclopentene, heptyne

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Answer:

Approximately $-205.7\; \rm kJ$ .

Explanation:

This question can be solved using Hess's Law.

Start by considering: how can the first three reactions (with known $\Delta H$ values) be combined to produce the reaction $\rm CH_4\; (g) + 4\; \rm Cl_2\; (g) \to CCl_4\; (g) + 4\; HCl\; (g)$ ?

Here's one possible combination:

Include the first reaction once, without inverting.
Invert the second reaction and include it once.
Include the third reaction after multiplying all its coefficients by two.

In other words, if $(1)$ , $(2)$ , and $(3)$ denote the three reactions with know $\Delta H$ values, respectively, then $1 \times (1) - 1 \times (2) + 2\times (3)$ will give the required reaction $\rm CH_4\; (g) + 4\; \rm Cl_2\; (g) \to CCl_4\; (g) + 4\; HCl\; (g)$ .

By Hess's Law, the $\Delta H$ value of the reaction $\rm CH_4\; (g) + 4\; \rm Cl_2\; (g) \to CCl_4\; (g) + 4\; HCl\; (g)$ will thus be:

$\begin{aligned}&1 \times \Delta H_1 - 1\times \Delta H_2 + 2\times \Delta H_3\\ &= 1 \times 74.6\; \rm kJ - 1 \times 95.7\; \rm kJ +2 \times (-92.3\; \rm kJ) \\ &= -205.7\; \rm kJ\end{aligned}$ .

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