for the reaction of hydrogen with iodine
H2(g) + I2(g) --> 2HI(g)
relate the rate of disappearance of hydrogen gas to the rate of formation of hydrogen iodide.
and my answer is:
rate formation of HI= delta [HI]/delta t
and the rate of reaction of I2 = delta [I2]/ delta t
to make 1/2 delta [HI]/ delta t = delta -[I2]/ delta t
That looks ok except the question is for H2 and not I2.
so change every I2 to H2
To relate the rate of disappearance of hydrogen gas to the rate of formation of hydrogen iodide, you can use the stoichiometry of the balanced chemical equation.
From the balanced equation:
H2(g) + I2(g) --> 2HI(g)
It is clear that 1 mol of hydrogen gas (H2) reacts with 1 mol of iodine gas (I2) to produce 2 moles of hydrogen iodide (2HI).
Therefore, the rate of disappearance of hydrogen gas (∆[H2]/∆t) can be related to the rate of formation of hydrogen iodide (∆[HI]/∆t) using the ratio of their coefficients.
In this case, the ratio is 1:2.
So, we can write:
(1/2) * (∆[HI]/∆t) = (∆[H2]/∆t)
This equation shows how the rates of hydrogen gas disappearance and hydrogen iodide formation are related.
To relate the rate of disappearance of hydrogen gas (H2) to the rate of formation of hydrogen iodide (HI), we need to consider the stoichiometry of the reaction. From the balanced equation:
H2(g) + I2(g) --> 2HI(g)
We can deduce that for every 1 mole of H2 that reacts, 2 moles of HI are formed. Therefore, the rate of formation of HI can be expressed as:
Rate of formation of HI = Δ[HI]/Δt
Similarly, we can express the rate of reaction of I2 as:
Rate of reaction of I2 = Δ[I2]/Δt
Now, based on the stoichiometry, we can say that for every 1 mole of HI formed, 1/2 mole of H2 is consumed. Hence, we can relate the rates of disappearance of H2 and formation of HI by noting that the rate at which H2 disappears is half of the rate at which HI is formed. Mathematically, we have:
(1/2) * (Δ[H2]/Δt) = Δ[HI]/Δt
So, to relate the rates, we divide the rate of the disappearance of H2 by 2 to obtain the rate of formation of HI.