Bengals QB Joe Burrow today was named AFC Offensive Player of the Week for Week 13 for his performance in the team's 27-24 win over the Kansas City Chiefs on Sunday.
Burrow completed 25 of 31 passes (80.6 percent) for 286 yards and two TDs for a passer rating of 126.6. His completion percentage tied for the second-highest in the NFL in Week 13, while his passer rating and 9.2 yards per attempt led all AFC quarterbacks. Burrow also ran 11 times for 46 yards with a four-yard TD on the game's opening drive. The performance marked the eighth time in Burrow's career and fifth time this season that he recorded a passing TD and rushing TD in the same game. His 46 rushing yards were one shy of tying a career high.
Burrow was particularly effective in the fourth quarter, which Cincinnati entered trailing 24-17. He completed nine of 11 passes for 80 yards in the quarter, including an eight-yard TD pass to HB Chris Evans with 8:54 left that proved to be the game-winner. When the Bengals got the ball back leading 27-24 at the 3:19 mark, Burrow engineered a game-clinching drive in which he went three for three and converted two third downs.
This is Burrow's fourth career AFC Offensive Player of the Week award. He won it in Week 4 of the 2021 season vs. Jacksonville, Week 16 of '21 vs. Baltimore, and Week 7 of this season vs. Atlanta. Bengals players have combined to win five weekly awards this year, with Burrow's two recognitions joining DE Trey Hendrickson (Defensive, Week 3), K Evan McPherson (Special Teams, Week 4) and HB Joe Mixon (Offensive, Week 9).
