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Calculate K at 298 K for the following reaction:

2NO(g) + Cl2(g) → 2NOCl(g)

Express answer in scientific notation.

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Alyssa Mae Remarim
Alyssa Mae RemarimLv10
26 Nov 2020

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Related questions

Consider the following equilibrium: 2NOCl(g) 2NO(g) + Cl2(g) with K = 1.6 x 10

The reaction 2NO (g) + Cl2 (g) -> 2NOCl (g) was studied at 363 K.

The following initial rates were obtained:

Experiment [NO] [Cl2] Rate of Reaction

1 . 0.01 0.01 70

2 0.01 0.02 140

3 0.02 0.02 140

What is the rate constant (k) for this reaction?

Calculate K at 298 K for the following reactions. Delta Gf kj/mol =

NOg...... 86.60

NO2g........... 51

O2.....0

Part A : NO(g) + 1/2 O2(g)......>> NO2(g) (answer in scientific notation, I am pretty sure the answer is 1.7 x 10^6)

Part B) Use the equilibrium constant ( K= 3.89 x 10^-34 at 298K ) to calculate Delta G (in kJ) at 298 for the following reaction:

2HCL (g)....> H2 (g) + CL2 (g)

answer delta G in kJ and in scientific notation please........

2HCL (g)....> H2 (g) + CL2 (g)

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