CH4+(x)O2=CO2 +(Y)H2O
C=1 +H=4 +O=? = C=1 +O=2+? +H=?
H=4>>Y=2
C=1 +H=4 +O=? = C=1 +O=(2+2) +H=4
C=1 +H=4 +O=4 = C=1 +O=4 +H=4
O=4>>X=2
CH4+(2)O2 =CO2 +(2)H2O
In this reaction 50% of the compound decompose in 10.5 min thus, it is half life of the reaction and denoted by symbol 
.
(a) For first order reaction, rate constant and half life time are related to each other as follows:
Thus, rate constant of the reaction is 
.
(b) Rate equation for first order reaction is as follows:
![k=\frac{2.303}{t_{1/2}}log\frac{[A_{0}]}{[A_{t}]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2.303%7D%7Bt_%7B1%2F2%7D%7Dlog%5Cfrac%7B%5BA_%7B0%7D%5D%7D%7B%5BA_%7Bt%7D%5D%7D)
now, 75% of the compound is decomposed, if initial concentration ![[A_{0} ]](https://tex.z-dn.net/?f=%5BA_%7B0%7D%20%5D)
is 100 then concentration at time t ![[A_{t} ]](https://tex.z-dn.net/?f=%5BA_%7Bt%7D%20%5D)
will be 100-75=25.
Putting the values,
On rearranging,
Thus, time required for 75% decomposition is 21 min.
There are approximately 160 grams in 1 mol of fe203 molecules therefore there would be 79/160 = 0.49375 mols of fe203 molecules in 79 grams therefore 5 atoms in total for each molecule of fe203 therefore 79/160 *5 =79/32=2.46875 mols in atoms
Answer:
<h2>Pressure will increase</h2>
Explanation:
At a constant temperature, the pressure of gas will increase proportional to the decrease in volume of the gas.
P1V1= P2V2
Decrease in volume result in increase in pressure as the equation has to hold true.
The balanced equation for the above reaction is as follows;
2S + 3O₂ --> 2SO₃
Stoichiometry of O₂ to SO₃ is 3:2
O₂ is the limiting reactant and S is provided in excess. since O₂ is the limiting reactant, the whole amount is consumed in the reaction and amount of product formed depends on amount of limiting reactant present.
Number of O₂ moles reacted- 4 g / 32 g/mol = 0.125 mol
3 mol of O₂ forms 2 mol of SO₃
therefore when 0.125 mol of O₂ reacts number of SO₃ moles - 2/3 x 0.125 mol
Number of SO₃ moles formed - 0.0833 mol
Answer is 4) 0.08 mol
