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Byju's Answer

Standard XIII

Chemistry

First Order Reaction

Ag → 2Bg + Cg...

Question

$A (g) \to 2 B (g) + C (g)$
Initially at t = 0 gas A was present along with some amount of gas C. At At t = 0, the mole fraction of gas C is $\frac{1}{3}$ . After some time $t = t_{1}$ , the total pressure is half of the final total pressure at $t = t_{x}$ (a very long time). Assume this decomposition to follow first order kinetics (at a constant temperature). It is also given at $t = t_{x}$ , the final total pressure is 35 bar.

At $t = t_{1}$ pressure of gas B is.

2.5 bar

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1.25 bar

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5.0 bar

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Data is insufficient

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Solution

The correct option is A 2.5 bar
$A (g) \to 2 B (g) + C (g)$
$\frac{2}{3} P_{i} 0 \frac{1}{3} P_{i}$ at $t = 0$
$\frac{2}{3} P_{i} - p 2 p \frac{1}{3} P_{i} + p$ at $t_{1}$
$\frac{2}{3} P_{i} - p_{x} 2 p_{x} \frac{1}{3} P_{i} + p_{x}$ at $t_{x}$
$P_{i} + 2 p = \frac{35}{2}$
$P_{i} + 2 p_{x} = 35$
after a very long time
$\frac{2}{3} P_{i} - p_{x} = 0$
by solving we get
$P_{i} = 15$
$2 p = 2.5$
pressure of $B$ at time t is 2.5 bar

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Similar questions

A (g) \to 2 B (g) + C (g)

Initially at t = 0 gas A was present along with some amount of gas C. At At t = 0, the mole fraction of gas C is

\frac{1}{3}

. After some time

t = t_{1}

, the total pressure is half of the final total pressure at

t = t_{x}

(a very long time). Assume this decomposition to follow first order kinetics (at a constant temperature). It is also given at

t = t_{x}

, the final total pressure is 35 bar.

At

t = t_{1}

pressure of gas B is.

A (g) \to 2 B (g) + C (g)

Initially at t = 0 gas A was present along with some amount of gas C. At At t = 0, the mole fraction of gas C is

\frac{1}{3}

. After some time

t = t_{1}

, the total pressure is half of the final total pressure at

t = t_{x}

(a very long time). Assume this decomposition to follow first order kinetics (at a constant temperature). It is also given at

t = t_{x}

, the final total pressure is 35 bar.

Ratio of rate constant at t = 0 to

t = t_{1}

t = t_{x}

is:

A (g) \to 2 B (g) + C (g)

Initially at t = 0 gas A was present along with some amount of gas C. At At t = 0, the mole fraction of gas C is

\frac{1}{3}

. After some time

t = t_{1}

, the total pressure is half of the final total pressure at

t = t_{x}

(a very long time). Assume this decomposition to follow first order kinetics (at a constant temperature). It is also given at

t = t_{x}

, the final total pressure is 35 bar.

Rate constant

(k) = (log 64 - log 49) s^{- 1}

. Value of

t_{1}

in seconds is:

Q. For a first order gas phase reaction:

A_{(g)} \to 2 B_{(g)} + C_{(g)}

P_{0}

be initial pressure of

A

and

P

, the total pressure at time '

t^{'}

. Integrated rate equation is:

Q. A certain mass of a gas A is kept in a closed container where it undergoes dimerisation, according to the reaction:

2 A (g) \to A_{2} (g)

Assuming that the temperature is constant, it was found that the partial pressure of

A_{2}

gas after time t was one-fifth of the initial pressure in the container. If each molecule of gas A weighs

10^{- 22} g

then select the correct statement(s):

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The correct option is A 2.5 barA(g)→2B(g)+C(g) 23Pi 0 13Pi at t=0 23Pi−p 2p 13Pi+p at t1 23Pi−px 2px 13Pi+px at tx Pi+2p=352 Pi+2px=35 after a very long time 23Pi−px=0 by solving we get Pi=15 2p=2.5 pressure of B at time t is 2.5 bar