A rigid stainless steel chamber contains 170 Torr of methane, CH4, and excess oxygen, O2, at 170.0 °C. A spark is ignited inside the chamber, completely combusting the methane. What is the change in total pressure within the chamber following the reaction? Assume a constant temperature throughout the process.

The change in total pressure is 0. There is no change in pressure.

Explanation:

Step 1: The balanced equation

CH4 + 2O2 → CO2 + 2H20

Step 2: Data given

Temperature = 170 °C = 443 Kelvin

170 Torr of methane, CH4, and excess oxygen, O2

Step 3: Calcuate change of pressure

For 1 mole of CH4 consumed, we need 2 moles of O2. This means:

140 torr of CH4 means the pressure of O2 is at least 2*140 = 280 torr. Since it's excess it will be more so let's consider 280 + x torr

For 1 mole of CH4 consumed, we need 2 moles of O2, to consume 1 mole of CO2 and 2 moles of H2O. We can consider 140 torr of CO2 and 280 torr of H2O. Since we are at 170 °C, the water produced is vapor.

Of the excess O2, there will, after 280 torr is used, remain x torr of O2.

280 torr of O2 is used.

From the Ideal Gas Law: PV = nRT (all parameters are constant)

That means, P1 = P2

Thus, change in total pressure = 0 (no change)