Question

question 6 (1 point)
in the chemical reaction below, the rate of consumption of oxygen is observed to be 4.0 mol/l·min. what is the rate of production of h₂o(g)?
ch₄(g) + 2 o₂(g) → co₂(g) + 2 h₂o(g)
4.0 mol/l·min
2.0 mol/l·min
1.0 mol/l·min
8.0 mol/l·min

Explanation:

Step1: Analyze stoichiometry

From the reaction $\ce{CH_{4(g)} + 2 O_{2(g)} -> CO_{2(g)} + 2 H_{2}O_{(g)}}$, the mole ratio of $\ce{O2}$ to $\ce{H2O}$ is $2:2$ (or $1:1$ when simplified).

Step2: Relate reaction rates

The rate of consumption of $\ce{O2}$ and rate of production of $\ce{H2O}$ are related by their stoichiometric coefficients. For every 2 moles of $\ce{O2}$ consumed, 2 moles of $\ce{H2O}$ are produced. So the rate of production of $\ce{H2O}$ should be equal to the rate of consumption of $\ce{O2}$ in this case? Wait, no, wait. Wait the coefficient of $\ce{O2}$ is 2, coefficient of $\ce{H2O}$ is 2. So the rate relationship is $\frac{\text{rate of consumption of } \ce{O2}}{2} = \frac{\text{rate of production of } \ce{H2O}}{2}$. So they are equal? Wait no, wait the rate of reaction is defined as $-\frac{\Delta[\ce{O2}]}{2\Delta t}=\frac{\Delta[\ce{H2O}]}{2\Delta t}$. So $-\frac{\Delta[\ce{O2}]}{\Delta t}=\frac{\Delta[\ce{H2O}]}{\Delta t}$. So if the rate of consumption of $\ce{O2}$ is $4.0\ \text{mol/L·min}$, then the rate of production of $\ce{H2O}$ is also $4.0\ \text{mol/L·min}$? Wait but that seems conflicting with the options? Wait no, wait maybe I made a mistake. Wait the coefficient of $\ce{O2}$ is 2, coefficient of $\ce{H2O}$ is 2. So the ratio of rate of $\ce{O2}$ consumption to rate of $\ce{H2O}$ production is $2:2 = 1:1$. So if $\ce{O2}$ is consumed at $4.0\ \text{mol/L·min}$, then $\ce{H2O}$ is produced at $4.0\ \text{mol/L·min}$? But let's check the options. One of the options is 4.0 mol/L·min. Wait but let's re-express the rate law. The rate of reaction $r = -\frac{1}{2}\frac{d[\ce{O2}]}{dt}=\frac{1}{2}\frac{d[\ce{H2O}]}{dt}$. So $-\frac{d[\ce{O2}]}{dt}=\frac{d[\ce{H2O}]}{dt}$. So if $-\frac{d[\ce{O2}]}{dt} = 4.0\ \text{mol/L·min}$, then $\frac{d[\ce{H2O}]}{dt}=4.0\ \text{mol/L·min}$. So the rate of production of $\ce{H2O}$ is $4.0\ \text{mol/L·min}$. Wait but that's one of the options. Wait but maybe I messed up. Wait no, let's take an example. Suppose in 1 minute, 4 moles of $\ce{O2}$ are consumed per liter. Then according to the reaction, 4 moles of $\ce{H2O}$ are produced per liter (since 2 moles $\ce{O2}$ produce 2 moles $\ce{H2O}$, so 4 moles $\ce{O2}$ produce 4 moles $\ce{H2O}$). So the rate is 4.0 mol/L·min.

Answer:

4.0 mol/L·min (the option with this value, e.g., if the first option is 4.0 mol/L·min, then the answer is that option)