How would you balance the complete combustion of acetylene by oxygen gas in a constant-pressure calorimeter?

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How would you balance the complete combustion of acetylene by oxygen gas in a constant-pressure calorimeter?

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Answer 1 · 2016-12-26T07:22:03

The basic equation is

$C_{2}$ $C_2$ $H_{2}$ $H_2$ + $O_{2}$ $O_2$ -----> C $O_{2}$ $O_2$ + $H_{2}$ $H_2$ O

The equation should be balanced as per the law of conservation of matter.

To balance the number of Hydrogen atoms on both the sides of the equation , I will add a $H_{2}$ $H_2$ O molecule on the Right Hand Side of equation.

$C_{2}$ $C_2$ $H_{2}$ $H_2$ + $O_{2}$ $O_2$ -----> C $O_{2}$ $O_2$ + 2 $H_{2}$ $H_2$ O

To balance the number of Carbon atoms on both the sides of the equation , I will add a $C_{2}$ $C_2$ $H_{2}$ $H_2$ molecule on the left Hand Side of equation and three molecules of C $O_{2}$ $O_2$ on the Right Hand side.

2 $C_{2}$ $C_2$ $H_{2}$ $H_2$ + $O_{2}$ $O_2$ -----> 4C $O_{2}$ $O_2$ + 2 $H_{2}$ $H_2$ O

To balance the number of Oxygen atoms on both the sides of the equation , I will add 4 molecules of $O_{2}$ $O_2$ on left hand side of the equation.

2 $C_{2}$ $C_2$ $H_{2}$ $H_2$ + 5 $O_{2}$ $O_2$ -----> 4C $O_{2}$ $O_2$ + 2 $H_{2}$ $H_2$ O

Answer 2 · 2016-12-26T09:18:12

Here's how I would approach it. To start off, the unbalanced equation was:

$C_{2} H_{2} (g) + O_{2} (g) \to {CO}_{2} (g) + H_{2} O (g)$ $"C"_2"H"_2(g) + "O"_2(g) -> "CO"_2(g) + "H"_2"O"(g)$

The hydrogens are balanced to begin with, so instinctively I would first balance the carbons by doubling the quantity of ${CO}_{2}$ $"CO"_2$ :

$C_{2} H_{2} (g) + O_{2} (g) \to 2 {CO}_{2} (g) + H_{2} O (g)$ $"C"_2"H"_2(g) + "O"_2(g) -> color(red)(2)"CO"_2(g) + "H"_2"O"(g)$

That changes the number of oxygens on the products side so that it is $2 \times 2 + 1 = 5 \times O$ $2xx2 + 1 = 5xx"O"$ . Now, nothing says we cannot use fractions to balance, just as long as we scale up the reaction to use whole numbers later.

(Fractional stoichiometries are OK, but are more conventionally used in enthalpy of formation reactions.)

So, since there are two oxygen atoms in one $O_{2}$ $"O"_2$ molecule, and there exist $5 \times O$ $5xx"O"$ atoms on the products side, let's say we have $\frac{5}{2} \times O_{2}$ $5/2xx"O"_2$ :

$C_{2} H_{2} (g) + \frac{5}{2} O_{2} (g) \to 2 {CO}_{2} (g) + H_{2} O (g)$ $"C"_2"H"_2(g) + color(red)(5/2)"O"_2(g) -> color(red)(2)"CO"_2(g) + "H"_2"O"(g)$

You can verify that there are these numbers of atoms on each side:

$stackrel("Reactants Side")overbrace(2xx"C")$ vs. $stackrel("Products Side")overbrace(2xx"C")$

$stackrel("Reactants Side")overbrace(5/2xx2xx"O")$ vs. $stackrel("Products Side")overbrace(2xx2xx"O" + 1xx"O")$

$stackrel("Reactants Side")overbrace(2xx"H")$ vs. $stackrel("Products Side")overbrace(2xx"H")$

So, technically, we balanced the equation. Now, just double everything to get rid of the fraction.

$color(blue)(2"C"_2"H"_2(g) + 5"O"_2(g) -> 4"CO"_2(g) + 2"H"_2"O"(g))$

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How would you balance the complete combustion of acetylene by oxygen gas in a constant-pressure calorimeter?

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