Question

learn by doing
calculate \\( \delta h^{\circ} \\) for the following reaction:
\\ 2 \mathrm{c}(s)+\mathrm{h}_{2}(g) \
ightarrow \mathrm{c}_{2} \mathrm{h}_{2}(g) \\
from the following information:
\\ \

$$\begin{array}{ll}\\operatorname{rxn}(\\mathrm{i}) \\quad \\mathrm{c}(s)+\\mathrm{o}_{2}(g) \ ightarrow \\mathrm{co}_{2}(g) & \\delta \\mathrm{h}^{\\circ}=-394 \\mathrm{~kj} \\\\ \\operatorname{rxn}(\\mathrm{ii}) \\mathrm{h}_{2}(g)+\\frac{1}{2} \\mathrm{o}_{2}(g) \ ightarrow \\mathrm{h}_{2} \\mathrm{o}(g) & \\delta \\mathrm{h}^{\\circ}=-286 \\mathrm{~kj} \\\\ \\operatorname{rxn}(\\mathrm{iii}) 2 \\mathrm{c}_{2} \\mathrm{h}_{2}(g)+5 \\mathrm{o}_{2}(g) \ ightarrow 4 \\mathrm{co}_{2}(g)+2 \\mathrm{h}_{2} \\mathrm{o}(l) & \\delta \\mathrm{h}^{\\circ}=-2599 \\mathrm{~kj}\\end{array}$$

\\
how does rxn (i) need to be manipulated?
reverse equation and multiply by 2
reverse equation
no change needed
reverse equation and multiply by 1 / 2
multiply by 2
multiply by 1 / 2
what is the resulting value for \\( \delta h \\) for rxn(i)?
\\( \delta h \\)
kj

Explanation:

Step1: Analyze the target reaction and rxn(i)

The target reaction has \(2C(s)\) on the left - hand side. Rxn(i) has \(C(s)\) on the left - hand side. To get \(2C(s)\) in the target reaction, we need to multiply rxn(i) by 2.

The original rxn(i): \(C(s)+O_{2}(g)\to CO_{2}(g)\), \(\Delta H^{\circ}=- 394\space kJ\)

After multiplying by 2: \(2C(s)+2O_{2}(g)\to 2CO_{2}(g)\)

According to Hess's law, when we multiply a reaction by a factor \(n\), the \(\Delta H\) of the reaction is also multiplied by \(n\).

Step2: Calculate the new \(\Delta H\) for the manipulated rxn(i)

If \(n = 2\) and the original \(\Delta H=-394\space kJ\), then the new \(\Delta H\) for \(2C(s)+2O_{2}(g)\to 2CO_{2}(g)\) is \(\Delta H=2\times(- 394)\space kJ=-788\space kJ\)

Answer:

For the first question (how to manipulate rxn(i)): multiply by 2.
For the second question (resulting \(\Delta H\) value for rxn(i)): \(-788\)