A sample of carbon is placed in a rigid 1.50 L flask. Dinitrogen monoxide is added and the flask heated to 800C. When the system reached equilibrium the partial pressure of the carbon dioxide is found to be 0.030 atm and the partial pressure of the dinitrogen monoxide to be 1.48 atm. What is the value of the equilibrium constant, Kp for the system at 800C?

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A sample of carbon is placed in a rigid 1.50 L flask. Dinitrogen monoxide is added and the flask heated to 800C. When the system reached equilibrium the partial pressure of the carbon dioxide is found to be 0.030 atm and the partial pressure of the dinitrogen monoxide to be 1.48 atm. What is the value of the equilibrium constant, Kp for the system at 800C?

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Answer 1 · 2014-05-22T18:45:36

The value of $K_{P}$ $K_P$ is 4.9 × 10⁻⁵.

1. Write the balanced chemical equation for the equilibrium.

C(s) + 2N₂O(g) ⇌ CO₂(g) + 2N₂(g)

2. Write the $K_{P}$ $K_P$ expression.

$K_{P} = \frac{P_{C O_{2}} P_{N_{2}}^{2}}{P_{N_{2} O}^{2}}$ $K_P = (P_(CO_2)P_(N_2)^2)/P_(N_2O)^2$

3 List the equilibrium concentrations.

$P_{C O_{2}}$ $P_(CO_2)$ = 0.030 atm

$P_{N_{2} O}$ $P_(N_2O)$ = 1.48 atm

There was no $N_{2}$ $"N"_2$ in the system at the beginning. All the $N_{2}$ $"N"_2$ present must have come from the reaction.

According to the equation, the reaction forms twice as much $N_{2}$ $"N"_2$ as ${CO}_{2}$ $"CO"_2$ . Thus,

$P_{N_{2}} = 2 P_{C O_{2}}$ $P_(N_2) = 2P_(CO_2)$ = 2 × 0.030 atm = 0.060 atm

4. Insert these values into the $K_{P}$ $K_P$ expression

$K_{P} = \frac{P_{C O_{2}} P_{N_{2}}^{2}}{P_{N_{2} O}^{2}} = \frac{0.030 \times {0.060}^{2}}{{(1.48)}^{2}}$ $K_P = (P_(CO_2)P_(N_2)^2)/P_(N_2O)^2 = (0.030 × 0.060^2)/(1.48)^2$ = 4.9 × 10⁻⁵

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