What is the increase in entropy when 11.2 l of o2 are mixed with 11.2 l of h2 at stp?
To calculate the increase in entropy when oxygen (O2) and hydrogen (H2) are mixed, we need to use the concept of ideal gases and the principles of thermodynamics. At standard temperature and pressure (STP), which is defined as 1 atmosphere (atm) and 273.15 Kelvin (K), both oxygen and hydrogen gases behave ideally. Therefore, we can use the ideal gas law and the concept of molar entropy to calculate the increase in entropy.
Step 1: Convert volume to moles
To calculate the moles of each gas, we need to use the ideal gas law equation:
PV = nRT
where
P = pressure (atm)
V = volume (L)
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (K)
Since the volume of both oxygen and hydrogen gases is 11.2 L, we can assume the pressure is 1 atm and the temperature is 273.15 K at STP. Therefore, the number of moles of oxygen (O2) and hydrogen (H2) can be calculated as follows:
For O2:
n(O2) = (P * V) / (R * T)
= (1 atm * 11.2 L) / (0.0821 L·atm/(mol·K) * 273.15 K)
For H2:
n(H2) = (P * V) / (R * T)
= (1 atm * 11.2 L) / (0.0821 L·atm/(mol·K) * 273.15 K)
Step 2: Calculate the initial entropy
The entropy increase can be obtained by calculating the difference in entropy between the final and initial states. Since both gases are initially in separate containers, we can calculate the initial entropy of the system by summing the entropies of the individual gases. The molar entropy (S) of each gas can be obtained from reference tables or using statistical thermodynamics.
For O2:
S(O2) = 205 J/(mol·K)
For H2:
S(H2) = 131 J/(mol·K)
The initial entropy (S_initial) can be calculated using the following equation:
S_initial = n(O2) * S(O2) + n(H2) * S(H2)
Step 3: Calculate the final entropy
When the gases are mixed, they react to form water vapor (H2O). The molar entropy of water vapor (S(H2O)) can be obtained from reference tables.
For H2O:
S(H2O) = 188 J/(mol·K)
Since the balanced chemical equation for the reaction is:
2H2(g) + O2(g) → 2H2O(g)
The number of moles of water vapor formed is equal to the lowest number of moles of the reactants, which is the moles of hydrogen (H2). Therefore, the number of moles of water vapor (n(H2O)) is equal to n(H2).
The final entropy (S_final) can be calculated using the following equation:
S_final = n(H2O) * S(H2O)
Step 4: Calculate the increase in entropy
The increase in entropy (∆S) can be obtained by subtracting the initial entropy from the final entropy:
∆S = S_final - S_initial
Finally, substitute the calculated values to obtain the final value of the increase in entropy.
To calculate the increase in entropy when 11.2 L of O2 are mixed with 11.2 L of H2 at STP (Standard Temperature and Pressure), we need to use the concept of entropy and the ideal gas law.
First, let's understand the concept of entropy. Entropy is a measure of the disorder or randomness of a system. When gases are mixed together, the overall disorder of the system increases, and therefore, the entropy increases.
To calculate the increase in entropy, we can use the equation:
ΔS = R * ln(Vf / Vi)
Where:
ΔS is the change in entropy
R is the ideal gas constant (8.314 J/(mol·K))
Vi is the initial volume
Vf is the final volume
In this case, the initial volume of O2 and H2 is 11.2 L each, and they are mixed. So the final volume will be the sum of the initial volumes (11.2 L + 11.2 L = 22.4 L).
Let's plug in the values into the equation:
ΔS = (8.314 J/(mol·K)) * ln(22.4 L / 11.2 L)
Simplifying the equation:
ΔS = 8.314 J/(mol·K) * ln(2)
Calculating the natural logarithm of 2:
ΔS ≈ 8.314 J/(mol·K) * 0.693
ΔS ≈ 5.76 J/(mol·K)
Therefore, the increase in entropy when 11.2 L of O2 is mixed with 11.2 L of H2 at STP is approximately 5.76 J/(mol·K).
Is the water gas or liquid. I wrote (g) but you should change it if necessary ans use the appropriate numbers.
2H2(g) + O2(g) => 2H2O(g)
What is the limiting reagent (LR)?
11.2 L H2 will require 5,6 L O2 and you have that much O2 so H2 is the LR.
Start ............11.2 L H2 + 11.2 L = 0 L H2O
change........-11.2..............-5.6 L.....11.2 L
equil................0..................5.6 L.....11.2 L
dSrxn = (n*So products) - (n*So reactants)
Post your work if you get stuck.