The force due to friction is generally independent of the contact area between the two surfaces. This implies regardless of whether you have two overwhelming objects of a similar mass, where one is half as long and twice as high as the other one, despite everything they experience the equivalent frictional power when you drag them over the ground.
Answer: The balanced equation for overall reaction is:
(CH3)3AuPH3 ----> C2H6 + (CH3)AuPH3
Explanation:
The reaction mechanism is given as follows:
Step 1: (CH3)3AuPH3 ↔ (CH3)3Au + PH3 (fast)
Step 2: (CH3)3Au → C2H6 + (CH3)Au (slow)
Step 3: (CH3)Au + PH3 → (CH3)AuPH3 (fast)
To balance this equation, firstly, we conduct proper atom count for each steps of the reaction mechanism.
It is important to note that for a reaction that involves several steps, the rate law is normally derived from the slow step ( which is step2 from the above mechanism).
Therefore, the balanced chemical equation for the overall reaction is:
(CH3)3AuPH3 ----> C2H6 + (CH3)AuPH3
Explanation:
Relation between entropy change and specific heat is as follows.

The given data is as follows.
mass = 500 g,
= 24.4 J/mol K
= 500 K,
= 250 K
Mass number of copper = 63.54 g /mol
Number of moles = 
= 
= 7.86 moles
Now, equating the entropy change for both the substances as follows.
= ![7.86 \times 24.4 \times [500 -T_{f}]](https://tex.z-dn.net/?f=7.86%20%5Ctimes%2024.4%20%5Ctimes%20%5B500%20-T_%7Bf%7D%5D)

= 750
So,
= 
- For the metal block A, change in entropy is as follows.

= ![24.4 log [\frac{375}{500}]](https://tex.z-dn.net/?f=24.4%20log%20%5B%5Cfrac%7B375%7D%7B500%7D%5D)
= -3.04 J/ K mol
- For the block B, change in entropy is as follows.

= ![24.4 log [\frac{375}{250}]](https://tex.z-dn.net/?f=24.4%20log%20%5B%5Cfrac%7B375%7D%7B250%7D%5D)
= 4.296 J/Kmol
And, total entropy change will be as follows.
= 4.296 + (-3.04)
= 1.256 J/Kmol
Thus, we can conclude that change in entropy of block A is -3.04 J/ K mol and change in entropy of block B is 4.296 J/Kmol.