JEE MAIN - Chemistry (2021 - 24th February Morning Shift - No. 17)
At 1990 K and 1 atm pressure, there are equal number of Cl2, molecules and Cl atoms in the
reaction mixture. The value of Kp for the reaction Cl2 (g) $$ \rightleftharpoons $$ 2Cl(g) under the above conditions
is x $$ \times $$ 10-1.
The value of x is _______. (Rounded off to the nearest integer)
The value of x is _______. (Rounded off to the nearest integer)
Answer
5
Explanation
Cl2(g) $$\rightleftharpoons$$ 2Cl(g)
Let moles of both of Cl2 and Cl molecule be x.
Partial pressure of Cl is, $${p_{Cl}} = {x \over {2x}} \times 1 = {1 \over 2}$$
Partial pressure of Cl2 is, $${p_{C{l_2}}} = {x \over {2x}} \times 1 = {1 \over 2}$$
Now, $${K_p} = {{{{({p_{Cl}})}^2}} \over {{p_{C{l_2}}}}} $$
$$\Rightarrow {K_p} = {{{{(1/2)}^2}} \over {1/2}} = {1 \over 2} = 0.5$$
= 5 $$\times$$ 10$$-$$1
Hence, x $$\times$$ 10$$-$$1
x = 5
Let moles of both of Cl2 and Cl molecule be x.
Partial pressure of Cl is, $${p_{Cl}} = {x \over {2x}} \times 1 = {1 \over 2}$$
Partial pressure of Cl2 is, $${p_{C{l_2}}} = {x \over {2x}} \times 1 = {1 \over 2}$$
Now, $${K_p} = {{{{({p_{Cl}})}^2}} \over {{p_{C{l_2}}}}} $$
$$\Rightarrow {K_p} = {{{{(1/2)}^2}} \over {1/2}} = {1 \over 2} = 0.5$$
= 5 $$\times$$ 10$$-$$1
Hence, x $$\times$$ 10$$-$$1
x = 5
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