Equilibrium Part 3

Chemistry Level 2

$2\text{CH}_4(g) \rightleftharpoons 3\text{H}_2(g)+ \text{C}_2\text{H}_2(g)$

The reaction above takes place in a rigid container.At $2000 \mathrm{K}$ , the reaction above has $K_c=.154$ .

If there are currently $2.00 \mathrm{M} \ \ \text{CH}_4$ and $.015 \mathrm{M} \ \ \text{C}_2\text{H}_2$ and $.010 \mathrm{M} \ \ \text{H}_2$ . What will happen to the pressure in the container as the reaction proceeds towards equilibrium given constant temperature throughout the reaction?

Note: $Q$ is the reaction quotient

The pressure will decrease since

Q<K_c

The pressure will increase since

Q>K_c

Can not be solved without initial pressures The pressure will increase since

Q<K_c

The pressure will remain unchanged The pressure will decrease since

Q>K_c

1 solution

Hamza A
Dec 27, 2018

$Q=\displaystyle \frac{[ \text{H}_2]^3_{i} [\text{C}_2\text{H}_2]_{i}}{[\text{CH}_4]^2_{i}}$

Where $i$ denotes initial and $[A]$ denotes the concentration of $A$ . Hence,

$Q=\displaystyle \frac{.010^3 \cdot .015}{2.00^2} =3.8 \times 10^{-9}$

Therefore we can deduce that $Q<K_c$ and the reaction will proceed to favor product formation. As the products have more reaction moles than the reactants $4_{\text{product}}>2_{\text{reactant}}$ the container's walls will experience more pressure as there are more moles of gas present. Therefore, the pressure will increase as $Q<K_c$

Equilibrium Part 3

1 solution

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