1A. 2NO(g) + 2H2(g) → N2(g) + 2H2O (g)
EXPERIMENT |
[NO]0 |
[H2]0 |
Initial rate (M/s) |
1 |
0.200 |
0.122 |
0.0313 |
2 |
0.400 |
0.122 |
0.1250 |
3 |
0.200 |
0.244 |
0.0625 |
Using the information above, find the order of each reactant and then find the rate constant.
1B. The decomposition of dinitrogen pentoxide happens by the following:
2N2O5 (g) → 4NO2(g) + O2(g)
Calculate the average rate (M/s) of decomposition given:
Time |
[N2O5] |
10 minutes |
1.238 x 10-2 |
20 minutes |
0.926 x 10-2 |
If this rate was constant What would the [N2O5] be at 1 hour?
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1A. 2NO(g) + 2H2(g) → N2(g) + 2H2O (g) EXPERIMENT [NO]0 [H2]0 Initial rate (M/s) 1...
The rate law for the reaction 2H2(g) + 2NO(g) ----> N2(g) + 2H2O(g) is rate=k[H2][NO]2. What is the rate constant at 800 degrees C when [NO] = 0.050 M and [H2] = 0.010 M and the rate of the reaction is 1.5 M/s
The first-order rate constant for the decomposition of N2O5, 2N2O5(g)→4NO2(g)+O2(g) at 70∘C is 6.82×10−3 s−1. Suppose we start with 2.40×10−2 mol of N2O5(g) in a volume of 2.1 L. a) How many moles of N2O5 will remain after 7.0 min? b) How many minutes will it take for the quantity of N2O5 to drop to 1.6×10−2 mol?
The first-order rate constant for the decomposition of N2O5, 2N2O5(g)→4NO2(g)+O2(g) at 70∘C is 6.82×10−3 s−1. Suppose we start with 2.30×10−2 mol of N2O5(g) in a volume of 1.5 L . a. How many moles of N2O5 will remain after 6 min ? b. How many minutes will it take for the quantity of N2O5 to drop to 1.9×10−2 mol ? c. What is the half-life of N2O5 at 70∘C?
The first-order rate constant for the decomposition of N2O5, 2N2O5(g)→4NO2(g)+O2(g) at 70∘C is 6.82×10−3 s−1. Suppose we start with 2.00×10−2 mol of N2O5(g) in a volume of 2.0 L . How many moles of N2O5 will remain after 7.0 min? How many minutes will it take for the quantity of N2O5 to drop to 1.6×10−2 mol? What is the half-life of N2O5 at 70∘C?
The first-order rate constant for the decomposition of N2O5, 2N2O5(g)→4NO2(g)+O2(g) at 70∘C is 6.82×10−3 s−1. Suppose we start with 2.60×10−2 mol of N2O5(g) in a volume of 2.3 L Part A How many moles of N2O5 will remain after 4.0 min ? Part B How many minutes will it take for the quantity of N2O5 to drop to 1.8×10−2 mol ? Part C What is the half-life of N2O5 at 70∘C?
The first-order rate constant for the decomposition of N2O5, 2N2O5(g)→4NO2(g)+O2(g) at 70∘C is 6.82×10−3 s−1. Suppose we start with 2.50×10−2 mol of N2O5(g) in a volume of 1.8 L . Part A: How many moles of N2O5 will remain after 4.0 min ? Part B: How many minutes will it take for the quantity of N2O5 to drop to 1.9×10−2 mol ? Part C: What is the half-life of N2O5 at 70∘C?
6) The rate constant for the first-order decomposition of N2O5 in the reaction 2N2O5(g) → 4NO2(g) + O2(g) is k=3.38 x 10-5 s-1 at 25°C. What is the half-life of N2O5? What will be the total pressure, initially 88.3 kPa for the pure N2O5 vapour, (a) 10 s, (b) 10 minutes after initiation of the reaction?
A proposed mechanism for the reaction 2NO(g) +2H2(g) → N2(g) + 2H2O(g): Step 1: 2NO(g) → N2O2(g) (very fast, reversible) Step 2: N2O2(g) + H2(g) → N2O(g) + H2O(g) (slow) Step 3: N2O(g) + H2(g) →N2(g) + H2O(g) (fast) What is the rate law for the overall reaction? O k[no]/2[Hz] O k[N20][H2] O k[NO]2 O k[NO]2[Hz] O k[NO]2[H212
For the reaction, 2NO(g)+2H2(g)→N2(g)+2H2O(g), what direction will the reaction proceed if [NO]=7.9x10-3M, [H2]=0.25 M, [N2]=0.15 M, & [H2O]=0.13 M. (K=650) forward reverse the reaction has stopped the reaction is at equilibrium
3. The reaction rate of the gas-phase reaction 2NO+2H2 → N2 + 2H2O was measured for several different initial pressures of the reactants. Determine the rate equation and thence the rate constant. [Hz] /mM [NO] /mM 26 20 26 d[N2]/dt /Ms? 6.17 x 10-S 1.57 x 10-5 7.59 x 10-5 1.14 104 10.1 26 26 19 28.5