4. The decomposition of phosphine, a very toxic gas, forms phosphorus and hydrogen in the following reaction:

4PH3 (g) P4 (g) + 6H2 (g)

(a) Express the rate of reaction with respect to each of the reactants and prducts
(b) If the instanteneous rate of the reaction with respect to PH3 is 0.34 M s-1, what is the instantaneous rate of the reaction
I ALREADY CALCULATED IT BUT ITS QUITE CONFUSING.WOULD U HELP ME WITH THIS.

a) Rate of rxn= (-1/4)(delta[PH3]/delta(t))=(delta[P4]/delta(t))=(1/6)(delta[H2]/delta(t))

b) Instantaneous rate of rxn= .085M/s

I'm trying to figure out how to post my answer explainations for this problem, but the website thinks that my explaination is a URL and won't allow it to be posted yet. I'm trying to get around it, if I do I will post in an additional comment!

What numbers did you get and how did you get them?

Of course, I can help you with this!

(a) To express the rate of reaction with respect to each of the reactants and products, we need to examine the stoichiometric coefficients in the balanced equation.

The stoichiometric coefficients represent the mole ratio between reactants and products. In this case, we have:

PH3: -4 (as it is consumed)
P4: +1 (as it is produced)
H2: +6 (as it is produced)

Therefore, the rate of reaction with respect to each component can be expressed as follows:

Rate of reaction with respect to PH3: -1/4 * (Δ[PH3] / Δt)
Rate of reaction with respect to P4: +1/1 * (Δ[P4] / Δt)
Rate of reaction with respect to H2: +1/6 * (Δ[H2] / Δt)

(b) If the instantaneous rate of the reaction with respect to PH3 is 0.34 M s-1, and we know that the stoichiometric coefficient for PH3 is -4, we can determine the instantaneous rate of the reaction as a whole:

Rate of reaction = -(1/4) * (Δ[PH3] / Δt)

Given that the instantaneous rate of the reaction with respect to PH3 is 0.34 M s-1, we can substitute this into the equation:

0.34 M s-1 = -(1/4) * (Δ[PH3] / Δt)

Now, we can solve for the instantaneous rate:

(Δ[PH3] / Δt) = -0.34 M s-1 * (-4)
(Δ[PH3] / Δt) = 1.36 M s-1

So, the instantaneous rate of the reaction as a whole is 1.36 M s-1.

Sure, I'd be happy to help you with this!

(a) To express the rate of reaction with respect to each reactant and product, we use the coefficients in the balanced chemical equation as exponents. Let's take a look at the equation again:

4PH3 (g) -> P4 (g) + 6H2 (g)

The rate of reaction with respect to phosphine (PH3) can be expressed as:

Rate = -(1/4) * Δ[PH3]/Δt

Note that the negative sign indicates the decrease in concentration of PH3 over time, since it is being consumed.

The rate of reaction with respect to phosphorus (P4) can be expressed as:

Rate = (1/1) * Δ[P4]/Δt

Since there is only one P4 molecule produced for every 4 PH3 molecules consumed, the coefficient is 1/4.

The rate of reaction with respect to hydrogen (H2) can be expressed as:

Rate = (1/6) * Δ[H2]/Δt

Since 6 H2 molecules are produced for every 4 PH3 molecules consumed, the coefficient is 1/6.

(b) If the instantaneous rate of the reaction with respect to PH3 is 0.34 M s^-1, we can use the rate expression for PH3 that we derived earlier:

Rate = -(1/4) * Δ[PH3]/Δt

Given that the instantaneous rate of PH3 is 0.34 M s^-1, we can substitute this value into the rate expression:

0.34 M s^-1 = -(1/4) * Δ[PH3]/Δt

Now we can rearrange the equation to solve for the instantaneous rate of the reaction, which is the rate of P4 and H2 formation:

Δ[PH3]/Δt = -0.34 M s^-1 * (-4)

Δ[PH3]/Δt = 1.36 M s^-1

So, the instantaneous rate of the reaction with respect to P4 and H2 formation is 1.36 M s^-1.

I hope this helps! Let me know if you have any further questions.