Question

8.0 mol of methane gas reacts completely in a 2.00L container containing excess O2 in 3.2s. Find average rate of consumption of methane

Answer 1

The average rate of consumption of methane in a reaction where 8.0 moles of methane are completely consumed in a 2.00 L container within 3.2 seconds is 1.25 mol/L/s.

Step-by-step explanation:

The average rate of consumption of methane during the given reaction can be calculated using the rate formula, which is the change in concentration of a reactant (or product) divided by the time over which the change occurs. Given that 8.0 moles of methane react completely in a 2.00 L container within 3.2 seconds, we can determine the rate of consumption.

First, we find the initial concentration of methane before the reaction which is a number of moles/volume. So we have 8.0 moles/2.00 L = 4.0 mol/L. Since the entire amount of methane is consumed within 3.2 seconds, the change in concentration is 4.0 mol/L. Therefore, the average rate at which methane is consumed is 4.0 mol/L divided by 3.2 s, which equals 1.25 mol/L/s. This is the average rate of consumption of methane in the reaction with oxygen.

Answer 2

1.3 M/s

Step-by-step explanation:

Step 1: Given data

• Initial amount of methane (ni): 8.0 mol

• Final amount of methane (nf): 0 mol (it reacts completely)

• Volume of the container (V): 2.00 L

• Time elapsed (t): 3.2 s

Step 2: Calculate the average rate of consumption of methane

Methane burns in excess oxygen according to the following equation.}

CH₄ + 2 O₂ ⇒ CO₂ + 2 H₂O

We can calculate the average rate of consumption of methane (r) using the following expression.

r = - Δn/V × t = - (nf- ni) / V × t

r = - (0 mol - 8.0 mol) / 2.00 L × 3.2 s = 1.3 M/s