Answer:
a) 0.11
b)76.9
c) 8.8
d) 1.7*10^-4
Explanation:
Step 1: Data given
K = 1.3 * 10^-2 for the reaction N2(g) + 3H2(g) ⇌ 2NH3(g)
Step 2: Formula of K
aA(g) + bB(g) ⇌ cC(g) + dD(g)
K = [C]^c *[D]^d / [A]^a * [B]^b
K = 1.3 * 10^-2 = [NH3]² / [H2]³*[N2]
Step 3:
a) 1/2N2 + 3/2H2(g) ⇌ NH3(g)
N2(g) + 3H2(g) ⇌ 2NH3
1/2N2 + 3/2H2(g) ⇌ NH3(g) =>K' = 
K' =
= 0.11
<em>b. 2NH3(g) ⇌ N2(g) + 3H2(g)</em>
N2(g) + 3H2(g) ⇌ 2NH3
2NH3(g) ⇌ N2(g) + 3H2(g) =>K' = 1/K
K' = 1/(1.3*10^-2) = 76.9
c. NH3(g) ⇌ 1/2 N2(g) + 3/2H2(g)
N2(g) + 3H2(g) ⇌ 2NH3
NH3(g) ⇌ 1/2 N2(g) + 3/2H2(g)
=>K' = 
K' = 
K' = 8.8
d. 2N2(g) + 6H2(g) ⇌ 4NH3(g)
N2(g) + 3H2(g) ⇌ 2NH3
2N2(g) + 6H2(g) ⇌ 4NH3(g)
K' = K²
K' = (1.3*10^-2)²
K' = 1.7 *10 ^-4
What classification system????
Answer:
The pressure of the gas is 0.278 atm
Explanation:
Step 1: Data given
Number of moles of the ideal gas = 0.540 moles
Volume = 35.5 L
Temperature = 223 K
Step 2: Calculate the pressure of the gas
p*V = n*R*T
⇒with p = the pressure of the gas = TO BE DETERMINED
⇒with V= the volume of the gas = 35.5 L
⇒with n = the moles of the gas = 0.540 moles
⇒with R = the gas constant = 0.08206 L*atm/mol*K
⇒with T = the temperature of the gas = 223 K
p = n*R*T / V
p = 0.540 * 0.08206 * 223 / 35.5
p = 0.278 atm
The pressure of the gas is 0.278 atm