For the following reaction. delta P(C_6 H_14) delta t was found to be - 3.1 time

Question

For the following reaction. delta P(C_6 H_14) delta t was found to be - 3.1 times 10^3 at m/s. C_6 H_14 (g) rightarrow C_6 H_6 (g) + 4H_2 (g) Determine delta P(H_2)/delta t for this reaction at the same time. 5.0E-2 a t m/s 1.2E-2 a t m/s 3.IE-3 a t m/s 1.9E-4 a t m/s 7.8E-4 a t m/s A chemical reaction of the general type A rightarrow 2B is first-order, with a rate constant of 1.91 times 10^-4 s^-1, Calculate the half-life of A and the time needed for the concentration of A to fall from 0.100) mol L^-1 to 0.050 mol L^-1. 5.24E3 s; 2.62E2 s 2.62E3 s; 2.62E3 s 3.63E3 s; 2.62E2 s 3.63E3 s; 3.63E3 s 5.24E3 s; 3.63E3 s The reaction CH_3 NC(g) rightarrow CH_3 CN(g) is first-order with respect to methyl isocyanide. CH_3 NC.If it takes 50.7 minutes for exactly 34.% of the initial amount of methyl isocyanide to react, w hat is the rate constant in units of min^-1? 6.7IE-3 min^-1 2.I3E-2 min^-1 1.30E-2 min^-1 I.02E-2 min^-1 8.20E-3 min^-1 Consider the reaction N_2 (g) + 0_2 (g) doubleheadarrow 2NO(g), for which K_c = 0.10 at 2,000 degree Starting with initial concentrations of0.09l M of N_2: and 0.091 M of O_2, determine the equilibrium concentration of NO. 6.67E-3 M 9.I0E-2 M 2.48E-2 M I) 8.68E-3 M I.24E-2M

Explanation / Answer

9)

rate of reaction can be written as:

rate = -dP(C6H14)/dt = 1/4 * dP(H2)/dt

So,

-dP(C6H14)/dt = 1/4 * dP(H2)/dt

3.1*10^-3 atm/s = 1/4* dP(H2)/dt

dP(H2)/dt = 1.2*10^-2 atm/s

Answer: B

I am allowed to answer only 1 question at a time

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.