Question

Write down the equilibrium condition for each of the following reactions:

$$\begin{gathered} \text{(a)}2\mathrm{H}\leftrightarrow {\mathrm{H}}_2 \\ \text{(b)}2\mathrm{CO}+{\mathrm{O}}_2\leftrightarrow 2{\mathrm{CO}}_2 \\ \text{(c)}{\mathrm{CH}}_4+2{\mathrm{O}}_2\leftrightarrow 2{\mathrm{H}}_2\mathrm{O}+{\mathrm{CO}}_2 \\ \text{(d)}{\mathrm{H}}_2{\mathrm{SO}}_4\leftrightarrow 2{\mathrm{H}}^{+}+{\mathrm{SO}}_4^{2-} \\ \text{(e)}2\mathrm{p}+2\mathrm{n}\leftrightarrow {}^4\mathrm{He} \end{gathered}$$

Short answer

Hence, the equilibrium condition for each reaction is given.

Step-by-step solution

Given information

We have the following reactions:

$$\begin{gathered} \text{(a)}2\mathrm{H}\leftrightarrow {\mathrm{H}}_2 \\ \text{(b)}2\mathrm{CO}+{\mathrm{O}}_2\leftrightarrow 2{\mathrm{CO}}_2 \\ \text{(c)}{\mathrm{CH}}_4+2{\mathrm{O}}_2\leftrightarrow 2{\mathrm{H}}_2\mathrm{O}+{\mathrm{CO}}_2 \\ \text{(d)}{\mathrm{H}}_2{\mathrm{SO}}_4\leftrightarrow 2{\mathrm{H}}^{+}+{\mathrm{SO}}_4^{2-} \\ \text{(e)}2\mathrm{p}+2\mathrm{n}\leftrightarrow {}^4\mathrm{He} \end{gathered}$$

Explanation

(a) For the first reaction, the equilibrium condition is:

$2{\mu }_{\mathrm{H}}=2{\mu }_{{\mathrm{H}}_2}$

(b) For the second reaction, the equilibrium condition is:

$2{\mu }_{\mathrm{CO}}+{\mu }_{{\mathrm{O}}_2}=2{\mu }_{{\mathrm{CO}}_2}$

(c) For the third reaction, the equilibrium condition is:

${\mu }_{{\mathrm{CH}}_4}+2{\mu }_{{\mathrm{O}}_2}=2{\mu }_{{\mathrm{H}}_2\mathrm{O}}+{\mu }_{{\mathrm{CO}}_2}$

(d) For the fourth reaction, the equilibrium condition is:

${\mu }_{{\mathrm{H}}_2{\mathrm{SO}}_4}=2{\mu }_{{\mathrm{H}}^{+}}+{\mu }_{{\mathrm{H}}_2{\mathrm{SO}}_4}$

(e) For the fifth reaction, the equilibrium condition is:

$2{\mu }_p+2{\mu }_n=4{\mu }_{\mathrm{He}}$