Answer:
Concentration, because the amounts of reactants and products remain constant after equilibrium is reached.
Explanation:
The rate of reaction refers to the amount of reactants converted or products formed per unit time.
As the reaction progresses, reactions are converted into products. This continues until equilibrium is attained in a closed system.
When equilibrium is attained, the rate of forward reaction is equal to the rate of reverse reaction, hence the concentration of reactants and products in the system remain fairly constant over time.
When deducing the rate of reaction, concentration of the specie of interest is plotted on the y-axis against time on the x-axis.
Answer:
3.41 x10⁶ torr
Explanation:
To solve this problem we need to remember the equivalency:
1 torr = 133.322 Pa
Then we can proceed to<u> convert 4.55×10⁸ Pa into torr.</u> To do that we just need to multiply that value by a fraction number, putting the unit that we want to convert <em>from</em> in the <em>denominator</em>, and the value we want to convert <em>to</em> in the <em>numerator</em>:
4.55x10⁸ Pa *
3.41 x10⁶ torr
1. big bang-the most accepted theory on the origin of the universe <span>
2. steady state-</span>all is the same and will always stay the same <span>
3. oscillating universe-</span>agrees with the big bang theory, but insists the universe expanded much quicker <span>
4. inflation-</span>it's like an inflating and deflating balloon that never stops
You must know the concentration of the acetic acid. Suppose the concentration is 0.1 M. The solution is as follows:
CH₃COOH → CH₃COO⁻ + H⁺
I 0.1 0 0
C -x +x +x
E 0.1 - x x x
Ka = (x)(x)/(0.1 - x)
1.8×10⁻⁵ = x²/(0.1 - x)
Solving for x,
x = 1.333×10⁻³ = H⁺
pH = -log[H⁺] = -log(1.333×10⁻³)
pH = 2.88
Answer:

Explanation:
Firstly, write the expression for the equilibrium constant of this reaction:
![K_{eq} = \frac{[ADP][Pi]}{ATP}](https://tex.z-dn.net/?f=K_%7Beq%7D%20%3D%20%5Cfrac%7B%5BADP%5D%5BPi%5D%7D%7BATP%7D)
Secondly, we may relate the change in Gibbs free energy to the equilibrium constant using the equation below:

From here, rearrange the equation to solve for K:

Now we know from the initial equation that:
![K_{eq} = \frac{[ADP][Pi]}{ATP}](https://tex.z-dn.net/?f=K_%7Beq%7D%20%3D%20%5Cfrac%7B%5BADP%5D%5BPi%5D%7D%7BATP%7D)
Let's express the ratio of ADP to ATP:
![\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BADP%5D%7D%7B%5BATP%5D%7D%20%3D%20%5Cfrac%7B%5BPi%5D%7D%7BK_%7Beq%7D%7D)
Substitute the expression for K:
![\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BADP%5D%7D%7B%5BATP%5D%7D%20%3D%20%5Cfrac%7B%5BPi%5D%7D%7BK_%7Beq%7D%7D%20%3D%20%5Cfrac%7B%5BPi%5D%7D%7Be%5E%7B-%5Cfrac%7B%5CDelta%20G%5Eo%7D%7BRT%7D%7D%7D)
Now we may use the values given to solve:
![\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}} = [Pi]e^{\frac{\Delta G^o}{RT}} = 1.0 M\cdot e^{\frac{-30 kJ/mol}{2.5 kJ/mol}} = 6.14\cdot 10^{-6}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BADP%5D%7D%7B%5BATP%5D%7D%20%3D%20%5Cfrac%7B%5BPi%5D%7D%7BK_%7Beq%7D%7D%20%3D%20%5Cfrac%7B%5BPi%5D%7D%7Be%5E%7B-%5Cfrac%7B%5CDelta%20G%5Eo%7D%7BRT%7D%7D%7D%20%3D%20%5BPi%5De%5E%7B%5Cfrac%7B%5CDelta%20G%5Eo%7D%7BRT%7D%7D%20%3D%201.0%20M%5Ccdot%20e%5E%7B%5Cfrac%7B-30%20kJ%2Fmol%7D%7B2.5%20kJ%2Fmol%7D%7D%20%3D%206.14%5Ccdot%2010%5E%7B-6%7D)