Question

A mixture of 0.3 mole of h2 and 0.3 mole of i2 is allowed to react in a 10 litre evacuated flask at 500oc , the reaction is h_2 +i_2\rightleftharpoons 2hi,k is found to be 64.The amount of unreacted i_2 at equilibrium is

Answer 1

Answer:

Explanation:

Kc=$HI^{2}$/$H^{2}$$I^{2}$,

64=$x^{2}$/0.03×0.03  

x2=64×9×10−4

x=8×3×10−2=0.24

x is the amount of HI at equilibrium amount of I2 at equilibrium will be 0.30−0.24=0.06

Answer 2

The equilibrium amount of unreacted iodine gas is 0.19 moles

Explanation:

We are given:

Initial moles of hydrogen gas = 0.3 moles

Initial moles of iodine gas = 0.3 moles

Volume of the flask = 10 L

Molarity is calculated by using the formula:

$\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume (in L)}}$

So, $\text{Initial molarity of hydrogen gas}=\frac{0.3}{10}=0.03M$

$\text{Initial molarity of Iodine gas}=\frac{0.3}{10}=0.03M$

The given chemical equation follows:

                     

Initial:        0.03    0.03

At eqllm:        0.03-x   0.03-x   2x

The expression of $K_c$ for above equation follows:

$K_c=\frac{[HI]^2]}{[H_2][I_2]}$

We are given:

$K_c=64$

Putting values in above expression, we get:

$64=\frac{2x}{(0.03-x)\times (0.03-x)}\\\\x=0.011,0.08$

Neglecting the value of x = 0.08 because the equilibrium concentration of hydrogen and iodine gas will become negative, which is not possible

So, concentration of unreacted iodine gas at equilibrium = (0.03 - x) = [0.03 - 0.011] = 0.019 M

$0.019M=\frac{\text{Equilibrium moles of iodine gas}}{10}\\\\\text{Equilibrium moles of iodine gas}(0.019\times 10)=0.19mol$

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