For the reaction C₆H₁₄(g) $\rarr$C₆H₆(g) + 4H₂(g), $\Delta$P(H₂)/$\Delta$t was found to be 2.5 * 10^-2 atm/s, where $\Delta$P(H₂) is the change in pressure of hydrogen. Determine $\Delta$P(C₆H₁₄)/$\Delta$t for this reaction at the same time.

Question

Quizplus · Accepted Answer

The stoichiometry of the reaction shows that 1 mole of C6H14 produces 4 moles of H2. Therefore, the rate of change of pressure for C6H14 is one-fourth that of H2, but negative because C6H14 is being consumed. Thus, $\Delta$P(C6H14)/$\Delta$t = -(1/4) * 2.5 * 10^-2 atm/s = -6.25 * 10^-3 atm/s, which is approximately -6.2 * 10^-3 atm/s.

For the Reaction C6H14(g) →\rarr→ C6H6(g) + 4H2(g) Δ\DeltaΔ P(H2) Δ\DeltaΔ T Was Found to Be 2

For the Reaction C₆H₁₄(g) $\rarr$ C₆H₆(g) + 4H₂(g) $\Delta$ P(H₂) $\Delta$ T Was Found to Be 2