Answer:
The correct option is: (D) -2.4 kJ/mol
Explanation:
<u>Chemical reaction involved</u>: 2PG ↔ PEP
Given: The standard Gibb's free energy change: ΔG° = +1.7 kJ/mol
Temperature: T = 37° C = 37 + 273.15 = 310.15 K (∵ 0°C = 273.15K)
Gas constant: R = 8.314 J/(K·mol) = 8.314 × 10⁻³ kJ/(K·mol) (∵ 1 kJ = 1000 J)
Reactant concentration: 2PG = 0.5 mM
Product concentration: PEP = 0.1 mM
Reaction quotient: ![Q_{r} =\frac{\left [ PEP \right ]}{\left [ 2PG \right ]} = \frac{0.1 mM}{0.5 mM} = 0.2](https://tex.z-dn.net/?f=Q_%7Br%7D%20%3D%5Cfrac%7B%5Cleft%20%5B%20PEP%20%5Cright%20%5D%7D%7B%5Cleft%20%5B%202PG%20%5Cright%20%5D%7D%20%3D%20%5Cfrac%7B0.1%20mM%7D%7B0.5%20mM%7D%20%3D%200.2)
<u>To find out the Gibb's free energy change at 37° C (310.15 K), we use the equation:</u>

![\Delta G = 1.7 kJ/mol + [2.303 \times (8.314 \times 10^{-3} kJ/(K.mol))\times (310.15 K)] log (0.2)](https://tex.z-dn.net/?f=%5CDelta%20G%20%3D%201.7%20kJ%2Fmol%20%2B%20%5B2.303%20%5Ctimes%20%288.314%20%5Ctimes%2010%5E%7B-3%7D%20kJ%2F%28K.mol%29%29%5Ctimes%20%28310.15%20K%29%5D%20log%20%280.2%29)
![\Delta G = 1.7 + [5.938] \times (-0.699) = 1.7 - 4.15 = (-2.45 kJ/mol)](https://tex.z-dn.net/?f=%5CDelta%20G%20%3D%201.7%20%2B%20%5B5.938%5D%20%5Ctimes%20%28-0.699%29%20%3D%201.7%20-%204.15%20%3D%20%28-2.45%20kJ%2Fmol%29)
<u>Therefore, the Gibb's free energy change at 37° C (310.15 K): </u><u>ΔG = (-2.45 kJ/mol)</u>
Answer:
B. temperature decreases as altitude increases.
Explanation:
Just like in the lower reaches of the atmosphere, the troposphere, in the mesosphere, temperature decreases as altitude increases.
The mesosphere is the third layer of the atmosphere just above the stratosphere.
- It begins at the top of the stratosphere and ends at the mesopause where the thermosphere begins.
- The mesosphere is often referred to as the middle layer.
With increasing height, the temperature of the mesosphere decreases significantly. The top of the mesosphere is one of the coldest part of the earth atmosphere. This is as a result of increasing atmospheric cooling by carbon dioxide in this region of the atmosphere.
Answer: The correct answer is A. 11.5 atm. The temperature is held constant at 293 K, therefore, we can use Boyle's Law to determine the initial pressure. Boyle's Law states that there is an inverse relationship between pressure and volume of gases. Therefore, as volume increases, the pressure will decrease and vice versa.
Further Explanation:
Boyle's Law can be mathematically expressed as:

In this problem, we are given the values:
P(initial) = ?
V(initial) = 80 L
P (final) = 0.46 atm
V (final) = 2000 L
Plugging in these values into the equation:

The initial pressure was 11.5 atm. Since the volume increased or expanded, the space where the gas particles move is bigger, so the frequency of collisions with the wall of the container and with other particles are effectively decreased. This, therefore, decreases the pressure from 11.5 to 0.46 atm.
Learn More
- Learn about Charles' Law brainly.com/question/1421697
- Learn about Ideal Gas Law brainly.com/question/6534668
- Learn about Gay - Lusaac's Law brainly.com/question/1358307
Keywords: gas, Boyle's Law, Ideal Gas Law