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Consider the reaction 4 extPH3( extg) ightarrow extP4( extg) + 6 extH2( extg). If extH2 is forming at a rate of 0.078 ext mol L⁻¹ exts⁻¹, what is the rate of formation of extP4 and the rate of react

Chemistry

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2 parts
a

0.013 mol L−1s−1\text{0.013 mol L}^{-1}\text{s}^{-1}\text{}

b

0.052 mol L−1s−1\text{0.052 mol L}^{-1}\text{s}^{-1}\text{}

Step-by-step solution

4 steps

You're right, Gezahagn! Let's finish up question 4.

4. Reaction: 4PH3(g)→P4(g)+6H2(g)4\text{PH}_3(\text{g}) \rightarrow \text{P}_4(\text{g}) + 6\text{H}_2(\text{g})
Given: Molecular hydrogen is being formed at the rate of 0.078 mol L−1s−10.078 \text{ mol L}^{-1}\text{s}^{-1}.
This means Δ[H2]Δt=0.078 mol L−1s−1\frac{\Delta[\text{H}_2]}{\Delta t} = 0.078 \text{ mol L}^{-1}\text{s}^{-1}.

Step 1: Calculate the overall reaction rate from the rate of formation of H2\text{H}_2.

Rate=16Δ[H2]Δt\text{Rate} = \frac{1}{6}\frac{\Delta[\text{H}_2]}{\Delta t} Rate=16×0.078 mol L−1s−1=0.013 mol L−1s−1\text{Rate} = \frac{1}{6} \times 0.078 \text{ mol L}^{-1}\text{s}^{-1} = 0.013 \text{ mol L}^{-1}\text{s}^{-1}

Part a· 2 steps

At what rate is P4\text{P}_4 being formed?

  1. Relate the rate of formation of P4\text{P}_4 to the overall reaction rate.

    The stoichiometric coefficient for P4\text{P}_4 is 1.

    Rate=11Δ[P4]Δt\text{Rate} = \frac{1}{1}\frac{\Delta[\text{P}_4]}{\Delta t}

  2. Calculate the rate of formation of P4\text{P}_4.

    Δ[P4]Δt=Rate=0.013 mol L−1s−1\frac{\Delta[\text{P}_4]}{\Delta t} = \text{Rate} = \boxed{\text{0.013 mol L}^{-1}\text{s}^{-1}\text{}}

Part b

0.052 mol L−1s−1\text{0.052 mol L}^{-1}\text{s}^{-1}\text{}

At what rate is PH3\text{PH}_3 reacting?

  1. Relate the rate of reaction of PH3\text{PH}_3 to the overall reaction rate.

    The stoichiometric coefficient for PH3\text{PH}_3 is 4.

    Rate=−14Δ[PH3]Δt\text{Rate} = -\frac{1}{4}\frac{\Delta[\text{PH}_3]}{\Delta t}

  2. Calculate the rate of reaction of PH3\text{PH}_3.

    −Δ[PH3]Δt=4×Rate=4×0.013 mol L−1s−1=0.052 mol L−1s−1-\frac{\Delta[\text{PH}_3]}{\Delta t} = 4 \times \text{Rate} = 4 \times 0.013 \text{ mol L}^{-1}\text{s}^{-1} = \boxed{\text{0.052 mol L}^{-1}\text{s}^{-1}\text{}}
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