Part A A certain first-order reaction has a rate constant...

Part A A certain first-order reaction has a rate constant of 2.30 !! 10?² s?¹ at 21 °C. What is the value of k at 70 °C if E? = 75.5 kJ?mol ? Express your answer in reciprocal seconds to three significant figures. k = [ ] s?¹ Part B Another first-order reaction also has a rate constant of 2.30 !! 10?² s?¹ at 21 °C. What is the value of k at 70 °C if E? = 124 kJ?mol ? Express your answer in reciprocal seconds to three significant figures. k = [ ] s?¹

Transcript

00:01 Hi, so we're calculating the other value of k at 70 degrees celsius given the activation energy and the other value of k at 21 degrees celsius.

00:13 Okay, so to solve this problem, we're going to use the two -point form of arrainius equation.

00:21 That's ln of k2 over k1, it's equivalent to activation energy over r multiplied by 1.

00:30 1 over t1 minus 1 over t 2 so this is the equation that we're going to use we will solve for k2 so ln of k2 and then k1 is 2 .30 times 10 to be negative to this is per second activation energy is 75 .5 this is in kilojoules per mole and then we need to convert this to jol.

01:01 Because r is in terms of joles one kilojole is a thousand joules okay divided by r this is the constant 8 .314 joles per mole k kelvin and the temperature t1 it's 21 degrees celsius so we need to add 2 .73 for this to be in kelvin minus 1 over 70 degrees celsius plus 273 this is also in kelvin so now we could cancel some minutes.

01:34 We'll be able to cancel kilojoules here and then joles and moles and the unit four temperature here.

01:41 So the remaining unit should be per second.

01:44 And calculating this one will have 4 .412 -5 -695.

01:53 This will be unit this.

01:53 And on the left side we still have ln of k2 over 2 .30 times 10 to the negative 2 per second.

02:01 So how do you? we remove l n we'll just use e s base erase eraser par of l n up k2 over 2 .30 times 10 to the negative 2 then here we also have e raised the power of 4 .4 .4126565 so we could down write this is k2 oops let me just rewrite this one k2 over 2 .30 times 10 to the negative 2 per second and then here we have e to the power of 4 .4 .1 .2 .65695.

02:43 That means k2.

02:45 Solving for this, we should get 1 .90 per second...

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