carbon monoxide (co) reacts with hydrogen (h₂) to form methane (ch₄) and water (h₂o). co(g) + 3h₂(g) ⇌…

carbon monoxide (co) reacts with hydrogen (h₂) to form methane (ch₄) and water (h₂o). co(g) + 3h₂(g) ⇌ ch₄(g) + h₂o(g) the reaction is at equilibrium at 1,000 k. the equilibrium constant of the reaction is 3.90. at equilibrium, the concentrations are as follows. co = 0.30 m h₂ = 0.10 m h₂o = 0.020 m what is the equilibrium concentration of ch₄ expressed in scientific notation? 0.0059 5.9×10⁻² 0.059 5.9×10²

carbon monoxide (co) reacts with hydrogen (h₂) to form methane (ch₄) and water (h₂o). co(g) + 3h₂(g) ⇌ ch₄(g) + h₂o(g) the reaction is at equilibrium at 1,000 k. the equilibrium constant of the reaction is 3.90. at equilibrium, the concentrations are as follows. co = 0.30 m h₂ = 0.10 m h₂o = 0.020 m what is the equilibrium concentration of ch₄ expressed in scientific notation? 0.0059 5.9×10⁻² 0.059 5.9×10²

Answer

Explanation:

Step1: Write equilibrium - constant expression

The equilibrium - constant expression for the reaction $CO(g)+3H_2(g)\rightleftharpoons CH_4(g)+H_2O(g)$ is $K = \frac{[CH_4][H_2O]}{[CO][H_2]^3}$.

Step2: Rearrange the expression to solve for $[CH_4]$

$[CH_4]=\frac{K[CO][H_2]^3}{[H_2O]}$.

Step3: Substitute the given values

Given $K = 3.90$, $[CO]=0.30\ M$, $[H_2]=0.10\ M$, and $[H_2O]=0.020\ M$. $[CH_4]=\frac{3.90\times0.30\times(0.10)^3}{0.020}$. First, calculate $(0.10)^3 = 0.001$. Then, $3.90\times0.30\times0.001=3.90\times0.0003 = 0.00117$. Finally, $\frac{0.00117}{0.020}=0.0585\approx0.059 = 5.9\times10^{-2}\ M$.

Answer:

B. $5.9\times 10^{-2}$