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melamori03 [73]
3 years ago
11

A sample of N2 gas is contaminated with a gas (A) of unknown molar mass. The partial pressure of each gas is known to be 200 tor

e at 25 degrees Celsius. The gas are allowed to Effuse through a pinhole, and it is found that has A escapes at 4 times the rate of N2. The molar mass of gas A is:
A. 1.75 g/mol
B. 448 g/mol
C. 112 g/mol
D. 7.01 g/mol
E. None of these
Chemistry
1 answer:
Tju [1.3M]3 years ago
7 0

Answer: The molar mass of gas A is 1.75 g/mol

Explanation:

To calculate the rate of diffusion of gas, we use Graham's Law.

This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation:

\text{Rate of effusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}

So,

\frac{\text{Rate}_A}{\text{Rate}_{N_2}}=\sqrt{\frac{M_{N_2}}{M_{A}}}

We are given:

Molar mass of nitrogen gas = 28 g/mol

Molar mass of A gas = ? g/mol

Putting values in above equation, we get:

\frac{\text{Rate}_A}{\text{Rate}_{N_2}}=\sqrt{\frac{28}{M_{A}}}\\4=\sqrt{\frac{28}{M_{A}}}

M_{A}=1.75g/mol

Thus molar mass of gas A is 1.75 g/mol

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The question is incomplete. The complete question is :

Hydrogen is manufactured on an industrial scale by this sequence of reactions:

$CH_3(g) + H_2O(g) \rightleftharpoons CO(g) + 3H_2(g    ) \ \ \ \ \ \ \ \ \ \ K_1$

$CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g) \ \ \ \ \ \ \ \ \ \ \ \  K_2$

The net reaction is  :

$CH_4(g) + 2H_2O(g) \rightleftharpoons CO_2(g) + 4H_2(g) \ \ \ \ \ \ \ \ \ K$

Write an equation that gives the overall equilibrium constant K in terms of the equilibrium constants K_1 and K_2. If you need to include any physical constants, be sure you use their standard symbols, which you'll find in the ALEKS Calculator.

Solution :

$CH_3(g) + H_2O(g) \rightleftharpoons CO(g) + 3H_2(g    ) \ \ \ \ \ \ \ \ \ \ K_1$

$K_1 = \frac{[CO][H_2]^3}{[CH_4][H_2O]}$     ...............(1)

$CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g) \ \ \ \ \ \ \ \ \ \ \ \  K_2$

$K_2 = \frac{[CO_2][H_2]}{[CO][H_2O]}$  ...................(2)

$CH_4(g) + 2H_2O(g) \rightleftharpoons CO_2(g) + 4H_2(g) \ \ \ \ \ \ \ \ \ K$

$K=\frac{[CO_2][H_2]^4}{[CH_4][H_2O]^2}$

On multiplication of equation (1) and (2), we get

$K_1 \times K_2=\frac{[CO][H_2]^3}{[CH_4][H_2O]} \times \frac{[CO_2][H_2]}{[CO][H_2O]}$

$K_1K_2=\frac{[CO_2][H_2]^4}{[CH_4][H_2O]^2}$  .................(4)

Comparing equation (3) and equation (4), we get

$K=K_1K_2$

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Answer:

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Enzymes generally have temperature and pH window for their activity.

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