Q13P - Chemistry: Molecular Nature Of Matter And Change

Question

Balance each reaction and write its reaction quotient, ${\mathbf{Q}}_{\mathbf{c}}$ :

$$\begin{gathered} \mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{ }\mathbf{ }{\mathbf{C}}_{\mathbf{2}}{\mathbf{H}}_{\mathbf{6}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{ }\mathbf{𝆏}\mathbf{ }\mathbf{ }{\mathbf{CO}}_{\mathbf{2}}\mathbf{ }\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{g}\mathbf{\right)} \\ \mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{ }{\mathbf{CH}}_{\mathbf{4}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{F}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{ }\mathbf{𝆏}\mathbf{ }\mathbf{ }{\mathbf{CF}}_{\mathbf{4}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{4}\mathbf{HF}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{,}\mathbf{ } \\ \mathbf{\left(}\mathbf{c}\mathbf{\right)}\mathbf{ }{\mathbf{SO}}_{\mathbf{3}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{ }\mathbf{𝆏}\mathbf{ }\mathbf{ }{\mathbf{SO}}_{\mathbf{2}}\mathbf{\left(}\mathbf{G}\mathbf{\right)}\mathbf{+}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{,}\mathbf{ } \end{gathered}$$

Step-by-Step Solution

Answer

The answer is,

$$\begin{gathered} \left(a\right) 2C_2{H}_6\left(g\right)+{70}_2\left(g\right) 𝆏 4C{O}_2 \left(g\right)+6H_{2}O\left(g\right),Q=\frac{{\left[C{F}_4\right]}^2{\left[H_{2}O\right]}^6}{{\left[C_2{H}_6\right]}^2{\left[O_2\right]}^7} \\ \left(b\right) C{H}_4\left(g\right)+4F_2\left(g\right) 𝆏 C{F}_4\left(g\right)+4HF\left(g\right), Q=\frac{\left[C{F}_4\right]{\left[HF\right]}^4}{\left[C{H}_4\right]{\left[F_2\right]}^4} \\ \left(c\right) 2S{O}_3\left(g\right) 𝆏 2S{O}_2\left(G\right)+O_2\left(g\right), Q=\frac{{\left[S{O}_2\right]}^2\left[O_2\right]}{{\left[S{O}_3\right]}^2} \end{gathered}$$

Step 1: Reaction quotient

The Reaction quotient can be written for equilibrium reactions. It is the ratio of concentrations of reactants and products. The reaction quotient is equal to the equilibrium constant for a reaction at equilibrium.

Step 2: Subpart (a)

The balanced chemical equation is shown below.

$2C_2{H}_6\left(g\right)+{70}_2\left(g\right) 𝆏 4C{O}_2 \left(g\right)+6H_{2}O\left(g\right)$

The reaction quotient is as shown below.

$Q=\frac{{\left[C{F}_4\right]}^2{\left[H_{2}O\right]}^6}{{\left[C_2{H}_6\right]}^2{\left[O_2\right]}^7}$

The reaction quotient is the ratio of the concentration of the reactants and products along with the stochiometry of the substances present in it.

Step 3: Subpart (b)

The balanced chemical equation is shown below.

$C{H}_4\left(g\right)+4F_2\left(g\right) 𝆏 C{F}_4\left(g\right)+4HF\left(g\right)$

The reaction quotient is as shown below.

$Q=\frac{\left[C{F}_4\right]{\left[HF\right]}^4}{\left[C{H}_4\right]{\left[F_2\right]}^4}$

The reaction quotient is the ratio of the concentration of the reactants and products along with the stochiometry of the substances present in it.

Step 4: Subpart (c)

The balanced chemical equation is shown below.

$2S{O}_3\left(g\right) 𝆏 2S{O}_2\left(G\right)+O_2\left(g\right)$\

The reaction quotient is as shown below.

$Q=\frac{{\left[S{O}_2\right]}^2\left[O_2\right]}{{\left[S{O}_3\right]}^2}$

The reaction quotient is the ratio of the concentration of the reactants and products along with the stochiometry of the substances present in it.