If C2H2 has an effusion rate of 0.0187moles/sec. What would the effusion rate for H2S be in moles/sec?

To solve this problem, we can use Graham's law of effusion, which states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. The formula for Graham's law is:

Rate1/Rate2 = sqrt(M2/M1)

Where:

• Rate1 and Rate2 are the rates of effusion of the two gases.
• M1 and M2 are the molar masses of the two gases.

Given:

• The rate of effusion of C2H2 (Rate1) is 0.0187 moles/sec.
• The molar mass of C2H2 (M1) is approximately 26 g/mol (122 + 12).
• The molar mass of H2S (M2) is approximately 34 g/mol (1*2 + 32).

We want to find the rate of effusion of H2S (Rate2).

Substituting the given values into the formula, we get:

0.0187/Rate2 = sqrt(34/26)

Solving for Rate2, we get:

Rate2 = 0.0187 / sqrt(34/26)

After calculating the above expression, we find that the rate of effusion of H2S is approximately 0.0142 moles/sec.