Question

Chemical Equilibrium and Chemical Kinetics

**Part A**

For a certain reaction, Kc = 8.85×1010 and kf= 7.52×10−2 M−2⋅s−1 . Calculate the value of the reverse rate constant, kr, given that the reverse reaction is of the same molecularity as the forward reaction. Express your answer with the appropriate units. Include explicit multiplication within units, for example to enter M−2⋅s−1 include ⋅ (multiplication dot) between each measurement.

**Part B**

For a different reaction, *K**c* =
1.70×10^{10}, *k*f=6.63×105s−1, and *k*r=
3.91×10^{−5} s−1 . Adding a catalyst increases the forward
rate constant to 2.06×10^{8} s−1 . What is the new value of
the reverse reaction constant, *k*r, after adding
catalyst?

Express your answer with the appropriate units. Include explicit
multiplication within units, for example to enter
*M*−2⋅*s*−1 include ⋅ (multiplication dot) between
each measurement.

**Part C**

Yet another reaction has an equilibrium constant
*K*c=4.32×105 at 25 ∘C. It is an exothermic reaction, giving
off quite a bit of heat while the reaction proceeds. If the
temperature is raised to200 ∘C , what will happen to the
equilibrium constant?

The equilibrium constant will

Yet another reaction has an equilibrium constant at 25 . It is an exothermic reaction, giving off quite a bit of heat while the reaction proceeds. If the temperature is raised to 200 , what will happen to the equilibrium constant?

A. | increase. |

B. | decrease. |

C. | not change. |

Answer #1

**A) we knwo that**

**Kc = Kf / Kr**

**so**

**8.85 x 10^10 = 7.52 x 10-2 / Kr**

**Kr = 8.5 x 10-13**

**so**

**kr value is 8.5 x 10-13 M-2 s-1**

**B)**

**while adding catalyst KC value remains same**

**so**

**1.70 x 10^10 = 2.06 x 10^8 / Kr**

**Kr = 0.0153**

**so**

**Kr value is 0.0153 M-2 s-1**

**C)**

**we know that**

**ln ( k2 / k1) = ( Ea /R) ln ( 1/ T1 - 1/T2)**

**now**

**T2 > T1**

**1 /T1 > 1 / T2**

**1 / T1 - 1/T2 > 0**

**also**

**for exothermic reaction Ea < 0**

**so**

**ln ( k2 / k1) < 0**

**so**

**k2 < k1**

**so**

**the equilibrium constant decreases**

**answer is**

**B) decrease**

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