2 HI(g) ⇄ H2(g) + I2(g) Kc = 0.0156 at 400ºC 0.550 moles of HI are placed in a 2.00 L container and the system is allowed to reach equilibrium. Calculate the concentration of H2 at equilibrium. 0.0275 M (Your correct answer) 0.138 M 0.275 M 0.550 M 0.220 M

The concentration of hydrogen gas at equilibrium is 0.0275 M

Explanation:

Molarity is calculated by using the equation:

Moles of HI = 0.550 moles

Volume of container = 2.00 L

For the given chemical equation:

Initial:                  0.275

At eqllm:           0.275-2x      x         x

The expression of for above equation follows:

We are given:

Putting values in above expression, we get:

Neglecting the negative value of 'x' because concentration cannot be negative

So, equilibrium concentration of hydrogen gas = x = 0.0275 M

Hence, the concentration of hydrogen gas at equilibrium is 0.0275 M

what are the choices?

i think the answer is terrae but not 100 percent sure. ▓▬▬ §