The given question is incomplete. The complete question is as follows.
Citrate synthase catalyzes the reaction
Oxaloacetate + acetyl-CoA 
citrate + HS-CoA
The standard free energy change for the reaction is -31.5 kJ*mol^-1
(
a) Calculate the equilibrium constant for this reaction a 37degrees C
Explanation:
(a). It is known that
, relation between change in free energy (
) of a reaction and equilibrium constant (K) is as follows.
where, T = temperature in Kelvin
The given data is as follows.
T = 310 K, 
(as 1 kJ = 1000 J)
Now, putting the given values into the above formula as follows.
ln K = 
= 
ln K = 12.22
K = antilog (12.22)
= 
Therefore, we can conclude that value of equilibrium constant for the given reaction is .
What this is??????????????????
Answer: The correct answer is option B i.e., 2.78 mol
Explanation:
Aluminium reacts with iodine to form Aluminium iodide
From the equation, it is clear that 3 moles of iodine reacts with 2 moles of Aluminium to form Aluminium iodide
We know 
For 2 moles of Aluminium, 
3 moles of Iodine reacts with 54 g of Aluminium
? moles of iodine react with 50 g of Aluminium
Answer: Theoretical Yield = 0.2952 g
Percentage Yield = 75.3%
Explanation:
Calculation of limiting reactant:
n-trans-cinnamic acid moles = (142mg/1000) / 148.16 = 9.584*10⁻⁴ mol
pyridium tribromide moles = (412mg/1000) / 319.82= 1.288*10⁻³ mol
- n-trans-cinnamic acid is the limiting reactant
The molar ratio according to the equation mentioned is equals to 1:1
The brominated product moles is also = 9.584*10⁻⁴ mol
Theoretical yield = (9.584*10⁻⁴ mol) * (Mr of brominated product)
= (9.584*10⁻⁴ mol) * (307.97) = 0.2952 g
Percentage Yield is : Actual Yield / Theoretical Yield = 0.2223/0.2952
= 75.3%
