consider the reaction below.\nh₂(g) + co₂(g) → h₂o(g) + co(g)\nat equilibrium at 600 k, the following are…

consider the reaction below.\nh₂(g) + co₂(g) → h₂o(g) + co(g)\nat equilibrium at 600 k, the following are true.\nco₂=9.5×10⁻⁴ m\nh₂=4.5×10⁻² m\nh₂o=4.6×10⁻³ m\nco=4.6×10⁻³ m\nwhat is the value of the equilibrium constant for this reaction in correct scientific notation?\n4.9×10⁻³\n4.9×10⁻²\n4.9×10⁻¹\n4.9×10³

consider the reaction below.\nh₂(g) + co₂(g) → h₂o(g) + co(g)\nat equilibrium at 600 k, the following are true.\nco₂=9.5×10⁻⁴ m\nh₂=4.5×10⁻² m\nh₂o=4.6×10⁻³ m\nco=4.6×10⁻³ m\nwhat is the value of the equilibrium constant for this reaction in correct scientific notation?\n4.9×10⁻³\n4.9×10⁻²\n4.9×10⁻¹\n4.9×10³

Answer

Explanation:

Step1: Write equilibrium - constant expression

For the reaction $H_2(g)+CO_2(g)\rightleftharpoons H_2O(g) + CO(g)$, the equilibrium - constant expression $K_c$ is given by $K_c=\frac{[H_2O][CO]}{[H_2][CO_2]}$.

Step2: Substitute the given concentrations

Substitute $[CO_2]=9.5\times 10^{-4}\ M$, $[H_2]=4.5\times 10^{-2}\ M$, $[H_2O]=4.6\times 10^{-3}\ M$, and $[CO]=4.6\times 10^{-3}\ M$ into the expression: [ \begin{align*} K_c&=\frac{(4.6\times 10^{-3})\times(4.6\times 10^{-3})}{(4.5\times 10^{-2})\times(9.5\times 10^{-4})}\ &=\frac{4.6\times4.6\times10^{-3 - 3}}{4.5\times9.5\times10^{-2-4}}\ &=\frac{21.16\times 10^{-6}}{42.75\times 10^{-6}} \end{align*} ]

Step3: Calculate the value of $K_c$

[ \begin{align*} K_c&=\frac{21.16}{42.75}\times\frac{10^{-6}}{10^{-6}}\ & = 0.495\times1\ &=4.95\times 10^{-1}\approx4.9\times 10^{-1} \end{align*} ]

Answer:

$4.9\times 10^{-1}$