The equation is 2 NH3 (g) ⇀↽ N2 (g) + 3 H2 (g)
Difference in the number of moles delta n = ((3 + 1) - 2) = 4 - 2 = 2
We have an equation Kp= Kc (R x T) ^ (delta n); R is constant and T = 300 K
Kp / Kc = (R x T) ^2 Based on the temperature value (300 K), we can conclude that Kp is Larger.
Answer:
The vapor pressure of benzaldehyde at 61.5 °C is 70691.73 torr.
Explanation:
- To solve this problem, we use Clausius Clapeyron equation: ln(P₁/P₂) = (ΔHvap / R) (1/T₁ - 1/T₂).
- The first case: P₁ = 1 atm = 760 torr and T₁ = 451.0 K.
- The second case: P₂ = <em>??? needed to be calculated</em> and T₂ = 61.5 °C = 334.5 K.
- ΔHvap = 48.8 KJ/mole = 48.8 x 10³ J/mole and R = 8.314 J/mole.K.
- Now, ln(P₁/P₂) = (ΔHvap / R) (1/T₁ - 1/T₂)
- ln(760 torr /P₂) = (48.8 x 10³ J/mole / 8.314 J/mole.K) (1/451 K - 1/334.5 K)
- ln(760 torr /P₂) = (5869.62) (-7.722 x 10⁻⁴) = -4.53.
- (760 torr /P₂) = 0.01075
- Then, P₂ = (760 torr) / (0.01075) = 70691.73 torr.
So, The vapor pressure of benzaldehyde at 61.5 °C is 70691.73 torr.
Answer:
Yes
Explanation: Had a question like this and I said yes and got it right
n = PV/RT
p = 1.6 atm
v = 12.7L
R = 0.0821
T = 24°C which is equivalent to 297.15 degrees k
n = (16 × 12.7) / (0.0821 × 297.15)
n = 20.32 / 24.39
n = 0.83 mol
C = 12.90
H = 1.0079
C2 = 12.010 × 2 = 24.02
H6 = 1.0079 × 6 = 6.0474
C2H6 = 30.0674
Ethane times n which is 30.0674 × 0.83mol
= 24.95 grams of C2H6. Which is Ethane.
