Lol i decided to make a science question for u guys to see if your good st science so here it is...:What is a decomposer,a produ
2 answers:
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Answer:
Explanation:
From the table, first find the order of reaction, then find the rate constant, write the rate equation, and, finally, subsititue the data for the reaction that starts with 0.45M of reagent A and 0.90 M of reagents B and C.
<u>1. Table</u>
Trial [A] (M) [B] (M) [C] (M) Initial rate (M/s)
1 0.20 0.20 0.20 6.0×10⁻⁵
2 0.20 0.20 0.60 1.8×10⁻⁴
3 0.40 0.20 0.20 2.4×10⁻⁴
4 0.40 0.40 0.20 2.4×10⁻⁴
<u>2. Orders</u>
a) From trials 3 and 4 you learn that the initial concentration of B does not change the change teh rate of the reaction. Hence the order with respect to [B] is 0.
b) From trials 1 and 2 you learn that when the concentration of C is tripled the rate of reaction is also tripled:
- 0.60 / 0.2 = 3, and
- 1.8×10⁻⁴ / 6.0×10⁻⁵ = 3
Hence, the order with respect to C is 1.
c) From trials 1 and 3 you get:
- 0.40/0.2 = 2
- 2.4×10⁻⁴ / 6.0×10⁻⁵ = 4
Which means that when the concentration of A is doubled, the rate of the reaction is quadruplicated. Hence, the order of the reaction with respect to A is 2.
<u>3. Rate equation</u>
Ther orders are:
Hence the rate is:
<u>4. Rate constant, k</u>
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You can use any trial to find the value of the constant, k
Using trial 1:
<u>5. Rate law:</u>
<u>6. Substitute</u>
Subsititue the data for the reaction that starts with 0.45M of reagent A and 0.90 M of reagents B and C.
Answer:
401135 kJ
Explanation:
From the balanced quation,
(q/n) = CΔE
Molar heat of combustion = 7.85kJk × (303.81-298.70)k
= 7.85kj × 5.11
= 40.1135kj
