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Question 8 The reaction rate expressions for the following reactions in terms of the disappearance of the reactants and the appearance of products: C$_6$H$_{14}$(g) $\rightarrow$ C$_6$H$_6$(g) + 4H$_2$(g) $\circ$ -$\Delta$[C$_6$H$_{14}$]/$\Delta$t = -$\Delta$[C$_6$H$_6$]/$\Delta$t = (1/4)$\Delta$[H$_2$]/$\Delta$t $\circ$ -$\Delta$[C$_6$H$_{14}$]/$\Delta$t = $\Delta$[C$_6$H$_6$]/$\Delta$t = (1/4)$\Delta$[H$_2$]/$\Delta$t $\circ$ -$\Delta$[C$_6$H$_{14}$]/$\Delta$t = -$\Delta$[C$_6$H$_6$]/$\Delta$t = -(1/4)$\Delta$[H$_2$]/$\Delta$t $\circ$ $\Delta$[C$_6$H$_{14}$]/$\Delta$t = $\Delta$[C$_6$H$_6$]/$\Delta$t = (1/4)$\Delta$[H$_2$]/$\Delta$t Moving to another question will save this response. 

          Question 8
The reaction rate expressions for the following reactions in terms of the disappearance of the reactants and the appearance of products:
C$_6$H$_{14}$(g) $\rightarrow$ C$_6$H$_6$(g) + 4H$_2$(g)
$\circ$ -$\Delta$[C$_6$H$_{14}$]/$\Delta$t = -$\Delta$[C$_6$H$_6$]/$\Delta$t = (1/4)$\Delta$[H$_2$]/$\Delta$t
$\circ$ -$\Delta$[C$_6$H$_{14}$]/$\Delta$t = $\Delta$[C$_6$H$_6$]/$\Delta$t = (1/4)$\Delta$[H$_2$]/$\Delta$t
$\circ$ -$\Delta$[C$_6$H$_{14}$]/$\Delta$t = -$\Delta$[C$_6$H$_6$]/$\Delta$t = -(1/4)$\Delta$[H$_2$]/$\Delta$t
$\circ$ $\Delta$[C$_6$H$_{14}$]/$\Delta$t = $\Delta$[C$_6$H$_6$]/$\Delta$t = (1/4)$\Delta$[H$_2$]/$\Delta$t
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Question 8 The reaction rate expressions for the following reactions in terms of the disappearance of the reactants and the appearance of products: C6H14(g) → C6H6(g) + 4H2(g) ∘ -Δ[C6H14]/Δt = -Δ[C6H6]/Δt = (1/4)Δ[H2]/Δt ∘ -Δ[C6H14]/Δt = Δ[C6H6]/Δt = (1/4)Δ[H2]/Δt ∘ -Δ[C6H14]/Δt = -Δ[C6H6]/Δt = -(1/4)Δ[H2]/Δt ∘ Δ[C6H14]/Δt = Δ[C6H6]/Δt = (1/4)Δ[H2]/Δt Moving to another question will save this response.

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Chemistry: Structure and Properties
Chemistry: Structure and Properties
Nivaldo Tro 2nd Edition
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The reaction rate expressions for the following reactions in terms of the disappearance of the reactants and the appearance of products C6H14(g) -> C6H6(g) + 4H2(g) -[C6H14]/t = -[C6H6]/t = -1/4[H2]/t 0 - [C6H14]/t = [C6H6]/t = 1/4[H2]/t [C6H14]/t = -[C6H6]/t = 1/4[H2]/t [C6H14]/t = [C6H6]/t = 1/4[H2]/t Moving to another question will save this response
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Transcript

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00:01 So we're writing the relative rate of reaction for each of these two reactions.
00:06 So we can figure out that our rate is the change in the concentration over time, the change in time.
00:18 And so we can just look at our coefficients and figure out how those rates are related then.
00:27 So we can see that for every two molecules of h2 that we lose, we also lose one molecule of o2, which allows us to gain two molecules of h2o.
00:40 So if i were to write the relative reaction, i'm just going to use those coefficients.
00:45 So the rate for h2 is twice the rate of o2, and it's going to be the same, or it's also going to be for water, it's going to be twice the rate of 02, meaning that the rate of disappearance of h2 and the rate of appearance of h2l are the same in magnitude...
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