14. Questions 14-17 refer to the following information. 2 PH,(g) 2 P(s) + 3 H2(g
ID: 1075162 • Letter: 1
Question
14. Questions 14-17 refer to the following information. 2 PH,(g) 2 P(s) + 3 H2(g) The thermal decomposition of phosphine (PH) into phosphorous and molecular hydrogen represented by the equation above is a first order process The experiment is run at 680°C in rigid container as the partial pressure of the phosphine gas is monitored. Two data points collected for the experiment are shown in the table below. Time (s) 0. 50. PPH3 (torr) 600. 300. Which of the following graphs would be a plot of the complete set of data for this reaction? 500 400 300 500 400 300 100 100 100 Time (s) 50 150 50 100 Time (s) 150 200 600 500 500 300 100 100 100 Time (s) 150 200 50 100 Time (s) 150 200Explanation / Answer
Solved the first three problems which are part to the same question, post multiple question to get the answer to Problem 17 as per Chegg guidelines
Q14)
2PH3(g) ----- 2P(s) + 3H2(g)
The reaction is first order, hence it will follows the first order kinetics
ln(Po/Pt) = kt
Since the pressure is getting to half of its original value (600) at t=50, hence 50s is the half-life of the reaction, so the pressure at 100 seconds will be
ln(600/Pt) = ln2/50 * 100
Pt = 150 torr
So the graph must be a decreasing graph (log style) and it must have a pressure of 150 torr at t=100
Therefore, the correct answer is Option C
Q15)
The pressure of PH3 will become zero
Now as per the reaction 2 moles of phosphine gas is giving rise to 3 moles of H2 gas
So 2n moles will give rise to 3n moles of H2 gas
P(H2)/P(PH3) = 3n/2n = 3/2
P(H2) = 3/2 * 600 = 900 torr
Hence the correct answer is Option D
Q16)
The correct answer is Option A
Since the reaction is first order wrt the only reactant, hence it is the slow phase of the reaction therefore the reaction must be a single step reaction
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