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?\no 4.9×10⁻³\no 4.9×10⁻²\no 4.9×10⁻¹\no 4.9×10³

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?\no 4.9×10⁻³\no 4.9×10⁻²\no 4.9×10⁻¹\no 4.9×10³

Answer

Explanation:

Step1: Write equilibrium - constant expression

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

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: $K_c=\frac{(4.6\times 10^{-3})\times(4.6\times 10^{-3})}{(4.5\times 10^{-2})\times(9.5\times 10^{-4})}$.

Step3: Calculate the numerator

$(4.6\times 10^{-3})\times(4.6\times 10^{-3}) = 4.6\times4.6\times10^{-3 - 3}=21.16\times 10^{-6}=2.116\times 10^{-5}$.

Step4: Calculate the denominator

$(4.5\times 10^{-2})\times(9.5\times 10^{-4})=4.5\times9.5\times10^{-2-4}=42.75\times 10^{-6}=4.275\times 10^{-5}$.

Step5: Calculate the equilibrium constant

$K_c=\frac{2.116\times 10^{-5}}{4.275\times 10^{-5}}=\frac{2.116}{4.275}\approx0.495\approx4.9\times 10^{-1}$.

Answer:

$4.9\times 10^{-1}$