D. With the same number of protons and different number of neutrons.
<span>In the field of science, usually, the product of an experiment is
computed ahead to understand if it reached a specific objective. It would reach
greater than 100% of percent yield if the factors include faster reaction rates;
proper handling of the reactants, no outside contaminants, and the procedure of
the experiment is followed smoothly. It would reach lesser than 100% percent yield
if the experiment is not followed, external factors such as contamination from
the environment (wind, moisture, etc). </span>
ΔG⁰ = ΔH⁰ - T ΔS⁰
ΔG⁰ : Standard free energy of formation of acetylene
ΔH⁰ : Standard enthalpy of formation (226.7 kJ/mol)
ΔS⁰ : Standard entropy change (58.8 J / K. mol)
T : Temperature 25°C = 298 K (room temperature)
ΔG⁰ = 226.7 - (298 x 58.8 x 10⁻³) = 209.2 kJ /mol
Answer:
(a) 7.11x10⁻⁴ M/s
(b) 2.56 mol.L⁻¹.h⁻¹
Explanation:
(a) The reaction is:
O₃(g) + NO(g) → O₂(g) + NO₂(g) (1)
The reaction rate of equation (1) is given by:
(2)
<u>We have:</u>
k: is the rate constant of reaction = 3.91x10⁶ M⁻¹.s⁻¹
[O₃]₀ = 2.35x10⁻⁶ M
[NO]₀ = 7.74x10⁻⁵ M
Hence, to find the inital reacion rate we will use equation (2):
Therefore, the inital reaction rate is 7.11x10⁻⁴ M/s
(b) The number of moles of NO₂(g) produced per hour per liter of air is:
t = 1 h
V = 1 L
![\frac{\Delta[NO_{2}]}{\Delta t} = rate](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%5BNO_%7B2%7D%5D%7D%7B%5CDelta%20t%7D%20%3D%20rate)
![\frac{\Delta[NO_{2}]}{\Delta t} = 7.11 \cdot 10^{-4} M/s*\frac{3600 s}{1 h} = 2.56 mol.L^{-1}.h{-1}](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%5BNO_%7B2%7D%5D%7D%7B%5CDelta%20t%7D%20%3D%207.11%20%5Ccdot%2010%5E%7B-4%7D%20M%2Fs%2A%5Cfrac%7B3600%20s%7D%7B1%20h%7D%20%3D%202.56%20mol.L%5E%7B-1%7D.h%7B-1%7D)
Hence, the number of moles of NO₂(g) produced per hour per liter of air is 2.56 mol.L⁻¹.h⁻¹
I hope it helps you!