JEE MAIN - Chemistry (2023 - 13th April Evening Shift - No. 13)
A(g) $$\to$$ 2B(g) + C(g) is a first order reaction. The initial pressure of the system was found to be 800 mm Hg which increased to 1600 mm Hg after 10 min. The total pressure of the system after 30 min will be _________ mm Hg. (Nearest integer)
Answer
2200
Explanation
At 10 minutes, the pressure increased by 2x:
$$800 + 2x = 1600$$
$$2x = 800$$
$$x = 400$$
The rate constant (k) can be found as:
$$k = \frac{2.303}{10} \log \frac{800}{400} = \frac{2.303 \times \log 2}{10}$$
For 30 minutes, we can set up the equation:
$$k = \frac{2.303}{30} \log \frac{800}{800 - y}$$
$$\frac{2.303 \times \log 2}{10} = \frac{2.303}{30} \log \left(\frac{800}{800 - y}\right)$$
Solving for y:
$$\left(\frac{800}{800 - y}\right) = 8$$
$$800 - y = 100$$
$$y = 700$$
Now we can find the total pressure after 30 minutes:
Total pressure = (800 - y) + 2y + y
Total pressure = 100 + 1400 + 700
Total pressure = 2200 mm Hg
So, the total pressure of the system after 30 minutes is 2200 mm Hg.
$$800 + 2x = 1600$$
$$2x = 800$$
$$x = 400$$
The rate constant (k) can be found as:
$$k = \frac{2.303}{10} \log \frac{800}{400} = \frac{2.303 \times \log 2}{10}$$
For 30 minutes, we can set up the equation:
$$k = \frac{2.303}{30} \log \frac{800}{800 - y}$$
$$\frac{2.303 \times \log 2}{10} = \frac{2.303}{30} \log \left(\frac{800}{800 - y}\right)$$
Solving for y:
$$\left(\frac{800}{800 - y}\right) = 8$$
$$800 - y = 100$$
$$y = 700$$
Now we can find the total pressure after 30 minutes:
Total pressure = (800 - y) + 2y + y
Total pressure = 100 + 1400 + 700
Total pressure = 2200 mm Hg
So, the total pressure of the system after 30 minutes is 2200 mm Hg.
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