Answer
(a)
$4 NO+O_2 \leftrightharpoons 2 N_2O_3$
$$Q_C = \frac{[Products]}{[Reactants]} = \frac{[ N_2O_3 ] ^{ 2 }}{[ NO ] ^{ 4 }[ O_2 ]}$$
(b)
$SF_6+ 2SO_3\leftrightharpoons 3 SO_2F_2$
$$Q_C = \frac{[Products]}{[Reactants]} = \frac{[ SO_2F_2 ] ^{ 3 }}{[ SF_6 ][ SO_3 ] ^{ 2 }}$$
(c)
$2SClF_5+H_2\leftrightharpoons S_2F_{10}+2HCl$
$$Q_C = \frac{[Products]}{[Reactants]} = \frac{[ S_2F_{10} ][ HCl ] ^{ 2 }}{[ SClF_5 ] ^{ 2 }[ H_2 ]}$$
Work Step by Step
(a)
- Balance the reaction.
$2 NO+O_2 \leftrightharpoons N_2O_3$
$2 NO+O_2 \leftrightharpoons 2N_2O_3$
$4 NO+O_2 \leftrightharpoons 2 N_2O_3$
- The exponent of each concentration is equal to its balance coefficient.
$$Q_C = \frac{[Products]}{[Reactants]} = \frac{[ N_2O_3 ] ^{ 2 }}{[ NO ] ^{ 4 }[ O_2 ]}$$
(b)
- Balance the reaction.
$SF_6+SO_3\leftrightharpoons SO_2F_2$
$SF_6+SO_3\leftrightharpoons 2 SO_2F_2$
$SF_6+ 2SO_3\leftrightharpoons 2 SO_2F_2$
$SF_6+ 2SO_3\leftrightharpoons 3 SO_2F_2$
- The exponent of each concentration is equal to its balance coefficient.
$$Q_C = \frac{[Products]}{[Reactants]} = \frac{[ SO_2F_2 ] ^{ 3 }}{[ SF_6 ][ SO_3 ] ^{ 2 }}$$
(c)
- Balance the reaction.
$SClF_5+H_2\leftrightharpoons S_2F_{10}+HCl$
$2SClF_5+H_2\leftrightharpoons S_2F_{10}+2HCl$
- The exponent of each concentration is equal to its balance coefficient.
$$Q_C = \frac{[Products]}{[Reactants]} = \frac{[ S_2F_{10} ][ HCl ] ^{ 2 }}{[ SClF_5 ] ^{ 2 }[ H_2 ]}$$
