Mahomes wins franchise record 6th offensive player of the week award

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KANSAS CITY, Mo. — Kansas City Chiefs quarterback Patrick Mahomes’ performance on Sunday in the victory over the Cleveland Browns earned him the AFC Offensive Player of the Week award.

Mahomes has now won the award a franchise record six times for a quarterback, passing Hall of Famer Joe Montana (5).

In the win against Cleveland, Mahomes went 27-for-36 and threw for 337 yards and three touchdowns. He also rushed for one touchdown.

Offensive player of the week – QB Patrick Mahomes – Kansas City Chiefs
Defensive player of the week – DE Maxx Crosby – Las Vegas Raiders
Special teams player of the week – K Evan McPherson – Cincinnati Bengals

Mahomes continued his September dominance in the win bringing his totals to 35 touchdowns, 0 interceptions and an 11-0 record in September.


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The Chiefs now point there attention to the upcoming road game against the Baltimore Ravens who are coming off an overtime loss to AFC West rivals the Las Vegas Raiders.

Mahomes is a perfect 3-0 against Lamar Jackson and the Ravens. The matchup will bring together some familiar faces.

Chiefs left tackle Orlando Brown joined the team from Baltimore in a trade during the offseason, and former Chiefs Justin Houston and Sammy Watkins will play their former squad.

Cornerback Marcus Peters (Chiefs 2015-17) has been sidelined by a torn ACL and will not play in Sunday’s game.

The two AFC heavyweights will kickoff from M&T Bank Stadium on Sunday, September 19 at 7:20 p.m. on Sunday Night Football.


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