Part A In order for a reaction to proceed spontaneously from left to right as written ________.
1 answer:
Answer:
d. The overall free-energy change must be negative
Explanation:
For a reaction to be spontaneous, The overall free-energy change must be negative for the reaction.
ΔG of a reaction can be calculated by subtracting the free energy of reactants from the free energy of products.
<u>Therefore, if a chemical reaction undergoes from a higher value of free energy (Reactants) to a lower value of free energy (Products), then ΔG will be negative in magnitude and the chemical reaction will be spontaneous. </u>
<u>Thus, ATP involved, Phosphate group transferred or a common intermediate must be formed or not, if free energy change is negative, the reaction will be spontaneous.</u>
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Answer: 0.596
Explanation:
For this problem, we want to find . To do so, we will need to use the ICE chart. The I in ICE is initial quantity. In this case, it is the initial pressure. Pressure is in atm. The C in ICE is change in each quantity. The E is equilibrium.
H₂(g) + I₂(g) ⇄ 2HI(g)
I 3.12 5.52 0
C -0.869 -0.869 +1.738
E 2.251 2.251 1.738
<u>For the steps below, refer to the ICE chart above.</u>
1. Since we were given the initial of H₂, I₂ and equilibrium of 2HI, we can fill those into the chart.
2. Since we were not given the initial for 2HI, we will put 0 in their place.
3. For the change, we need to add pressure to the products to make the reaction reach equilibrium. We would add on the products and subtract from the reactants to equalize the reaction. Since we don't know how much the change in, we can use variable x. We know that the equilibrium of 2HI is 1.738, we know the change is 0.869 because 1.738/2=0.869. Since there are 2 moles of HI, we must divide the equilibrium by 2 to find x, so that we can fill that into the reactants side.
4. With the equilibrium values, we can find the equilibrium pressure. It is products over reactants. We use the values of E.
The [HI]² comes from 2HI. We more the moles to the exponent when we are calculating the equilibrium pressure.
We typically don't put units because K is unitless, but know that the units is atm throughout the entire problem.