Express the rate of the reaction in terms of the change in concentration of each of the reactants and products.
2D(g)+3E(g)+F(g)>>>>2G(g) + H(g) when D is decreasing at 0.30 MOL/Ls, how fast is (H) increasing?
rate D is 1/2*d(D)/dT which is 1/2*change in (D)/time.
rate E is 1/2*d(E)/dT
rate F is d(F)/dT
etc.
rate = -d(D)/2*dT = -1/2*(0.3) = -0.15 M/s = rate
rate = d(H)/dT = 0.15 M/s
To express the rate of the reaction in terms of the change in concentration of each reactant and product, we can calculate the reaction rate using the stoichiometric coefficients.
Given the balanced chemical equation: 2D(g) + 3E(g) + F(g) → 2G(g) + H(g)
From the equation, we can see that for every 2 moles of D consumed, 1 mole of H is produced.
Using this information, we can write the rate equation:
Rate = -1/2 * (∆[D]/∆t)
Since D is decreasing at a rate of 0.30 MOL/Ls, we substitute ∆[D]/∆t with -0.30 MOL/Ls:
Rate = -1/2 * (-0.30 MOL/Ls)
Simplifying this expression, we find:
Rate = 0.15 MOL/Ls
Therefore, H is increasing at a rate of 0.15 MOL/Ls.
To express the rate of the reaction in terms of the change in concentration of each reactant and product, we need to use the coefficients from the balanced chemical equation. In this case, the balanced equation is:
2D(g) + 3E(g) + F(g) ⟶ 2G(g) + H(g)
The rate of the reaction can be expressed as:
Rate = Δ[H(g)] / Δt
where Δ[H(g)] is the change in concentration of H over a given time interval Δt.
From the balanced equation, we can see that the stoichiometric coefficient of H(g) is 1. This means that for every 2 moles of D that react, 1 mole of H is produced.
Given that D is decreasing at a rate of 0.30 MOL/Ls, this means that the rate of reaction for the consumption of D can be expressed as:
Rate of D consumption = -Δ[D(g)] / Δt = -0.30 MOL/Ls
Since the stoichiometric coefficient of D is 2, we can relate the rate of D consumption to the rate of H production as:
Rate of H production = (-1/2) * Rate of D consumption
= (-1/2) * (-0.30 MOL/Ls)
= 0.15 MOL/Ls
Therefore, the rate at which H is increasing is 0.15 MOL/Ls.