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

6.68%

Explanation:

FV= 1000

N = 25*2=50

PMT= 8.75%*1000/2= 43.75

PV = -1250

I/Y = RATE(25*2, 8.75%*1000/2, -1200, 1000)

I/Y = 0.033397

I/Y = 3.34%

YTM = 3.34% * 2

YTM = 6.68%