Question

The reaction, $CO(g)+3H_{2}(g) \rightleftharpoons CH_{4}(g)+H_{2}O(g)$, is at equilibrium at 1300 K in a 1L flask. It also contain 0.30 mol of CO, 0.10 mol of $H_{2}$ and 0.02 mol of $H_{2}O$ and an unknown amount of $CH_{4}$ in the flask. Determine the concentration of $CH_{4}$ in the mixture. The equilibrium constant, Kc, for the reaction at the given temperature is 3.90.

Answer 1

"Let the concentration of methane at equilibrium be x.

$C{ O }_{ g }\quad +\quad 3{ H }_{ 4(g) }\quad \leftrightarrow \quad C{ H }_{ 4(g) }\quad +\quad { H }_{ 2 }(g)$

At equilibrium,

$CO\quad \Rightarrow \quad \frac { 0.3 }{ 1 } \quad =\quad 0.3\quad M$

${ H }_{ 2 }\quad \Rightarrow \quad \frac { 0.1 }{ 1 } \quad =\quad 0.1\quad M$

${ H }_{ 2 }O\quad \Rightarrow \quad \frac { 0.02 }{ 1 } \quad =\quad 0.02\quad M$

It is given that,

${ K }_{ C } = 3.90$

$\frac { \left[ { CH }_{ 4(g) } \right] \left[ { H }_{ 2 }{ O }_{ g } \right]}{ \left[ { CO }_{ (g) } \right] { \left[ { H }_{ 2(g) } \right]}^{ 3 } } \quad =\quad { K }_{ C }$

${ K }_{ C }\quad =\quad \frac { x\quad \times \quad 0.02 }{ 0.3\quad \times \quad { (0.1) }^{ 3 } } \quad =\quad 3.90$

$x\quad =\quad \frac { 3.90\quad \times \quad 0.3\quad \times \quad { (0.1) }^{ 3 } }{ 0.02 } \quad =\quad \frac { 0.00117 }{ 0.02 } \quad =\quad 0.0585$

$M\quad =\quad 5.85\quad \times \quad { 10 }^{ -2 }\quad M$

Hence the concentration of ${ CH }_{ 4 } at \quad equilibrium\quad is\quad 5.85\quad \times \quad { 10 }^{ -2 }\quad M$"

Answer 2

HEY MATE....☺✌☺

${CH_4\:at \: equilibrium \:is\:5.85}$